finalreduc

1

EXPLICIT FORMS OF DISCONTINUOUS FUNCTIONS, THE DIRAC DELTA,

AND IRREDUCIBLE FORMS

M. KYLE MURPHY

Abstract. In this paper, we construct and prove explicit forms composed of closed-form expres-

sions of the Heaviside Step, Ramp, Signum, Absolute Value, and Dirac Delta functions. The

explicit forms of the first four functions lead to a discussion concerning their points of disconti-nuity, which result from a pair of non-removable, overlapping branch points. The explicit form of

the Dirac Delta function warrants the most analysis. We argue the Dirac Delta function‘s selectorproperty is responsible due to the complex residue about a pair of non-removable, overlapping

simple-poles instead of an infinitely-tall, infinitesimally-thin spike; additionally, from the explicit

form, we demonstrate the Dirac Delta has no sign. We also show the explicit form preserves allproperties of the Dirac Delta function. From the Dirac Delta follows our discourse upon irreducible

forms of functions necessary when the function‘s range includes a point of indeterminacy. We il-

lustrate the usefulness and necessity of irreducible forms through two applications, one of which isa direct and consistent calculation of the Laplacian of the inverse distance function only possible

through the use of an irreducible form.

1. Introduction

Considering real, discontinuous functions, we discuss the following five: the Heaviside Step, Ramp,Signum, Absolute Value, and Dirac Delta functions with purpose to mathematically and conceptu-ally clarify, most importantly, the Dirac Delta function. In his famous monograph upon QuantumMechanics, Dirac [7] introduced to his audience a Delta function to account for certain infinite quan-tities that emerged in his discussion upon the observables of a quantum system. He continues byclaiming the Delta function as an improper function, and as a tall, narrow spike at the origin toaccount for the selector property. Later work, most notably the theory of distributions, pioneeredby Schwartz, enabled mathematics a comfortable description regarding the Dirac Delta function. Athorough and satisfying history of the Dirac Delta function beginning with the work of Cauchy isgiven by Katz and Tall [11].

In our paper we prove the Dirac Delta function is, verily, a mathematical function; we give its ex-plicit, closed-form expression and elaborate upon its interesting singularity structure. Furthermore,we introduce two irreducible forms, one which appears in the explicit definition of the Dirac Deltafunction; the other enables us to directly arrive the result ∇2(1/r) = −4πδ3(~r) from the Laplacianof the inverse distance function 1/r. We begin the paper by constructing the explicit form of theHeaviside Step function with the proof of the following theorem.

Theorem 1.1. For z ∈ R, the Heaviside Step function, θ(z), takes the following explicit form:

θ(z) =1

2+ i

ln(z)− ln(−z)2π

.

We prove the explicit form by considering z > 0, z < 0 and discuss the point z = 0. Fromlogarithmic identities in [1] we show that, for z > 0, the explicit form evaluates to 1/2 + 1/2 = 1;for z < 0 the explicit form evaluates to 1/2 − 1/2 = 0. Work towards an explicit form of theHeaviside Step function has been conducted previous to our study. Venetis [12] gives an analyticform involving the summation of two inverse tangent functions, however the underlying singularitystructure is left ambiguous. From the explicit Heaviside Step function, we prove the explicit forms

1Keywords: Discontinuous functions, Heaviside step, Ramp, Signum, Absolute Value, Dirac Delta, explicit form,

irreducible forms.

1

2 M. KYLE MURPHY

of the Ramp, Signum, and Absolute Value functions with relative identities. Complex analysis isparamount in much of the analysis conducted within our paper. As evident in the explicit form ofthe Heaviside Step function, we have two branch points we discuss. These branch points are thesource of discontinuity; they are overlapping at the point z = 0, however non-removable as theyresult in a 2π jump between the real positives and the real negatives (the cut is a closed curve).Following, we derive the explicit form of the Dirac Delta function.

Theorem 1.2. The Dirac Delta function, δ(z), for z ∈ R has the following explicit form:

δ(z) =i

2π

(1

z− 1

z

).

We are able to arrive at the explicit form by differentiating the Heaviside Step function from therelation dθ(z)/dz = δ(z). We must also mention the strikingly strong parallels between M. Sato‘shyperfunction representations of the Heaviside Step and the Dirac Delta functions as given in [20],[10] and [9]. For every value of z ∈ R, except z = 0, the explicit form yields zero. When z = 0 wehave an indeterminate form: the location of discontinuity in the Dirac Delta function and the pointof selection when applied underneath integration. The emergence of 1/z − 1/z segues into Section8 concerning irreducible forms.

Lemma 1. Let z ∈ R, then the irreducible form of z/z appears asz

z= θ(z) + θ(−z).

We prove the irreducible form by first considering sgn(|z|) = |z|/||z|| = z/z. Through identities ofthe Signum function and the Heaviside Step function, we arrive at the irreducible form. Additionally,we note the derivative d/dz(θ(z) + θ(−z)) = 2δ(z), leading to a corollary proving that the DiracDelta function, similar to zero, has no sign.

Corollary 1. δ(z) = −δ(z).

By computing −δ(z) in explicit form, we show the fractions 1/z − 1/z absorb the negative sign,resulting in δ(z) = −δ(z). This allows the imaginary unit to be written in either the numerator orthe denominator. The sign neutrality of the Dirac Delta function is actually expected; the DiracDelta function is zero everywhere except at a singular point where it assumes an indeterminatevalue.

Lemma 2. Let z ∈ R, then the function 1/z − 1/z takes the irreducible form

1

z− 1

z= 2πiδ(z).

This is trivially shown from the explicit form of the Dirac Delta function. The last, main theoremof the paper concerns the general selector property of the Dirac Delta function and the residue aboutthe poles appearing in the form 1/z − 1/z.

Theorem 1.3. Let g(z) be a continuously differentiable function with non-degenerate roots zi ∈ Rand f(z) any function provided it is continuous at f(zi). For the general selector property of theDirac Delta function∫ ∞

−∞f(z)δ(g(z))dz =

∑i

Resz=zi

(f(z)

|g′(zi)|(z − zi)

)=∑i

f(zi)

|g′(zi)|,

where the summation is performed over the roots of g(z).

Following in the proof of this theorem is our demonstration that the selector property of theDirac Delta function arises from the complex residue about a pair of simple, overlapping (yet notremovable) poles instead of the thin and narrow spike representation intersecting the function f(z)at the sampling point. These overlapping poles originate from the overlapping branch points in the

EXPLICIT FORMS OF DISCONTINUOUS FUNCTIONS, THE DIRAC DELTA, AND IRREDUCIBLE FORMS 3

Heaviside Step function. The rest of the paper involves two applications exhibiting the usefulnessof irreducible forms and the multidimensional Dirac Delta function. The first application considersthe vorticity of a planar, irrotational vortex, providing a result of the vorticity proportional to thecirculation of the vortex and the 2-D polar Dirac Delta function. The second application solvesthe Laplacian of the inverse distance function 1/r with the inclusion of the irreducible form r/r =θ(r)+θ(−r). From the calculation of 1

r2∂∂r

(r2 ∂∂r

(1r

))we reach the conclusion ∇2(1/r) = −4πδ3(~r).

The appendices A, B, and C conclude our paper. Appendix A tabulates many of the existingproperties of the Dirac Delta function. Appendix B proves all the properties in Appendix A holdfor the explicit form of the Dirac Delta function. Lastly, Appendix C demonstrates the Fouriertransformations of the explicit forms presented evaluate to the Fourier transformations of theircounterparts. Appendic C also calculates the Fourier transformations of the two presented irreducibleforms.

2. Heaviside Step Function

Definition 1. The Heaviside Step function is a discontinuous function of a real variable with jumpdiscontinuity located at zero. It takes the constant value 0 for negative argument and 1 for positiveargument. The Heaviside Step function is denoted as θ(z) throughout this paper. It is prescribedthe following piecewise description [14]

(2.1) θ(z) =

0 if z < 0

Indeterminate if z = 0

1 if z > 0

.

We also have an identity of the Heaviside Step function regarding itself with negative argument as:θ(z) = 1− θ(−z).

Theorem 2.1. For z ∈ R, the Heaviside Step function, θ(z), takes the following explicit form:

(2.2) θ(z) =1

2+ i


.

Proof. Equation (2.2) has been constructed from common logarithmic identities given in [1], namelymaking use of ln(−z) = ln(z) + iπ. It possesses an equivalent image as the Heaviside Step function:0 for [−∞, 0); 1 for (0,∞]. We prove equation (2.2) as the Heaviside Step function via evaluationfor all z ∈ R. For z > 0

θ(z) =1

2+ i


=1

2+ i

ln(z)− (ln(z) + iπ)

2π

=1

2− i2π

2π= 1 .

For z < 0

θ(−z) =1

2+ i

ln(−z)− ln(−(−z))2π

=1

2+ i

(ln(z) + iπ)− ln(z)

2π

=1

2+i2π

2π= 0 .

We now evaluate equation (2.2) at z = 0, where we expect a jump discontinuity. Taking limits asthey approach zero from the negative and positive gives

limz→0−

1

2+ i


= 0,

4 M. KYLE MURPHY

and

limz→0+

1

2+ i


= 1.

As expected, the limit does not exist and there is a jump discontinuity from 0 to 1. Furthermore,calculating at z = 0 brings about the indeterminate form ∞−∞ due to the logarithms. As such,we have satisfied the piecewise definition of the Heaviside Step function (2.1). We make note certainapplications of θ(z) have a defining value for θ(0), however this is only an assigned convention. Weclose our proof by concluding (2.2)

θ(z) =1

2+ i


.

�

3. Ramp Function

Definition 2. The Ramp function is a discontinuous function of a real variable with point disconti-nuity located at zero. The Ramp function is the integral of the Heaviside Step function; for negativeargument, R(z) evaluates to a constant zero and for positive argument, R(z) is simply z. From [15],it is defined with the Heaviside Step function as follows

R(z) =

∫ z

−∞θ(z′)dz′(3.1)

= zθ(z),(3.2)

and can be written in the following piecewise description

(3.3) R(z) =

{0 if z < 0

z if z ≥ 0.

Theorem 3.1. For z ∈ R, the Ramp function, R(z), has the following explicit form:

(3.4) R(z) =z

2+ i

z(ln(z)− ln(−z))2π

.

Proof. We have a couple avenues in which to navigate this proof. First we shall consider the simplestidentity (3.2) in which to arrive at the explicit form (3.4). From (2.2)

R(z) = zθ(z)

=z

2+ i


.

Secondly, we use (3.1) to arrive at the explicit form of the Ramp function. Using (2.2) as the formof the Heaviside Step function

R(z) =

∫ z

−∞

1

2+ i

ln(z′)− ln(−z′)2π

dz′.(3.5)

Since the Heaviside Step function outputs zero for a negative argument, we may rewrite the boundsof (3.5) as

R(z) =

∫ z

0

1

2+ i

ln(z′)− ln(−z′)2π

dz′.(3.6)

Computing the integral (3.6) produces

z

2+ i


−[z′

2+ i

z′(ln(z′)− ln(−z′))2π

]z′=0

.


We cannot directly evaluate the lower bound of integration due to the emergence of the form 0(∞−∞). Therefore we take the limit as z′ → 0, leading us once again to

R(z) =z

2+ i


,

which agrees with the piecewise description (3.3). Thus we have proven (3.4)

R(z) =z

2+ i


.

�

4. Signum Function

Definition 3. The Signum function, or Sign function, is a function of a real variable with jumpdiscontinuity located at zero. The function returns the sign of its argument (a real number) attachedto unity, ergo its name. The jump discontinuity at z = 0 is from −1 to 1. It is described for z ∈ Rwith a piecewise as given by [17]

(4.1) sgn(z) =

−1 if z < 0

Indeterminate if z = 0

+1 if z > 0

.

The Signum function is defined with the Heaviside Step function by

(4.2) sgn(z) = 2θ(z)− 1,

from which it can be shown that

(4.3) sgn(z) = θ(z)− θ(−z).

Additionally, for real argument

(4.4) sgn(z) =z

|z|=|z|z.

Theorem 4.1. For z ∈ R, the Signum function, sgn(z), takes the following explicit form:

(4.5) sgn(z) = iln(z)− ln(−z)

π.

Proof. We begin the proof with relation (4.2)

sgn(z) = 2θ(z)− 1.

Using equation (2.2) as the Heaviside Step function, we are able to arrive at (4.5),

sgn(z) = 2

(1

2+ i


)− 1

= iln(z)− ln(−z)

π.

Computing (4.5) at z > 0, z < 0, and z = 0 demonstrates agreement with the piecewise definitionof the Signum function, (4.1). We then conclude

sgn(z) = iln(z)− ln(−z)

π.

�

6 M. KYLE MURPHY

5. Absolute Value Function

Definition 4. The Absolute Value function is a discontinuous function of a real variable with pointdiscontinuity at z = 0. The function is often thought of as a number’s distance from zero, andreturns a positive value for any argument not equal to zero. As by [16] it can be described with thefollowing piecewise

(5.1) |z| =

{z if z ≥ 0

−z if z < 0.

The absolute value function has the following relationships with the Signum fuction

(5.2) |z| = sgn(z)z,

and

(5.3)d|z|dz

= sgn(z).

Theorem 5.1. For z ∈ R, the absolute value function, |z|, has an explicit form:

(5.4) |z| = iz(ln(z)− ln(−z))

π.

Proof. We begin the proof with relation (5.2)

|z| = sgn(z)z.

This simply involves multiplying (4.5) by z, which gives

|z| = iz(ln(z)− ln(−z))

π.

Using (5.3), we recover the Signum function by taking the derivative of (5.4)

d|z|dz

= iln(z)− ln(−z)

π= sgn(z).

Considering (5.4) for all z ∈ R returns a positive z except for the case of z = 0. At z = 0 wearrive at the same indeterminate form encountered in the evaluation of the Ramp function at z = 0,that is 0(∞−∞), however both limits of (5.4) from the negative and positive sides approach zero,therefore we define |0| = 0. With this definition, (5.4) coincides with the given piecewise (5.1) forthe Absolute Value function; we have proven

|z| = iz(ln(z)− ln(−z))

π.

�

6. Overlapping Branch Points

Before we continue to the Dirac Delta function, we first analyze the class of singular pointswe have encountered so far. Doing so aids us to classify the singular point in the Dirac Deltafunction. The four functions we have worked with all have singularity z = 0; the Ramp andAbsolute Value functions are ascribed a defining value at these locations: R(0) = 0 and |0| = 0 dueto the existence of their limits. It is trivial to note these points for the Heaviside Step function andSignum function are jump discontinuities, however we hesitate to classify the discontinuities in theRamp and Absolute Value functions as removable discontinuities (in spite of our defining values forthese points). Expanding our singularity analysis to the complex plane illuminates the point z = 0.We investigate the function responsible for the singularities:

(6.1) f(z) = ln(z)− ln(−z),


by first examining the similar function

(6.2) fε(z) = ln(ε+ z)− ln(ε− z).

Note fε(z) has logarithmic branch points located at z = ±ε with branch cut [−ε, ε]. By lettingε→ 0 fε(z) becomes f(z) and the branch points overlap at z = 0. The branch cut diminishes untilε = 0 at which point the cut disappears and the thinning strip connecting the Riemann sheets issnipped off. Furthermore, we may alter the branch cut at will albeit all possible configurations ofthe cut are a closed curve in the complex plane with beginning and end branch points at z = 0. Thisis to say (6.1) has disconnected Riemann sheets. The point z = 0 of (6.1) is a pair of overlapping,non-self-removing logarithmic branch points. To show why the branch points do not automaticallyremove themselves we consider the piecewise definition of (6.1)

(6.3) ln(z)− ln(−z) =

−iπ if x > 0

iπ if x < 0

Indeterminate if x = 0

We match (6.3) with a legal branch cut representation beginning from the first branch point at(0 + 0i), extending northward to (0 +∞i), encompassing the positives to (∞+∞i), traveling downthe imaginaries to (∞−∞i), returning to real zero at (0−∞i), and closing the curve by attachingthe cut to the second branch point located at (0 + 0i). Thus the branch cut engulfs half of thecomplex plane and accounts for the 2π jump witnessed in (6.3). To say the overlapping branch cutsremove themselves would be to disregard the 2π jump and Riemann sheet structure of (6.1).

Remark. Due to certain properties relating the Dirac Delta function with the Heaviside Step function,this fact of overlapping branch cuts supports a later argument that the Dirac Delta function has apair of overlapping, non-self-removing simple poles located at z = 0.

7. Dirac Delta Function

Definition 5. The Dirac Delta function, δ(z), in the traditional sense, is not considered a function[7][3][13]. It is considered a generalized function or a measure in most modern mathematical litera-ture [13]. The Dirac Delta function has a domain of R; it equals 0 everywhere except at z = 0 whereit takes an indeterminate value, and can be described with the piecewise

(7.1) δ(z) =

{0 if z 6= 0

Indeterminate* if z = 0.

The Dirac Delta function is defined as the derivative of the Heaviside Step function

(7.2)dθ(z)

dz= δ(z).

There are many known properties of the Dirac Delta function: properties concerning itself, propertiesunder differentiation, and properties under integration. These are listed in Appendix A.

*Very often, the Dirac Delta is conceptualized to be an infinitely tall, infinitesimally thin spikeat z = 0, represented as ∞ at z = 0, such as its treatment in [13][7][3][19]. This definition isoften required to explain the Dirac Delta function’s selector property under the operation of in-tegration. In this section, we discover and prove the Dirac Delta function is not an infinitelytall, infinitesimally thin spike at z = 0, but instead a pair of simple, overlapping poles lo-cated at z = 0. Thus it is not a tall spike breaching a function’s curve at z = 0, but insteadthe residue about the poles at z = 0 in the Dirac Delta function which lends it its selector prop-erty. Additionally, the indeterminacy of the Dirac Delta function at z = 0 is delineated in the proofof its explicit form.

8 M. KYLE MURPHY

Theorem 7.1. The Dirac Delta function for z ∈ R has the following explicit form:

(7.3) δ(z) =i

2π

(1

z− 1

z

).

Proof. We begin with identity (7.2) by differentiating the explicit form of θ(z): equation (2.2)

dθ(z)

dz=

d

dz

(1

2+ i


)=

i

2π

d

dz(ln(z)− ln(−z))

δ(z) =i

2π

(1

z− 1

z

).

We now aim to produce the Heaviside Step function, (2.2), from δ(z) with the relation∫δ(z)dz =

θ(z). Indefinite integration gives∫δ(z)dz =

∫i

2π

(1

z− 1

z

)dz,

=i

2π(ln(z)− ln(z) + C).

We define C as a constant of integration. We let C = −iπ− iπ and make use of common logarithmicidentities given in [1] to arrive at (2.2)

i

2π(ln(z)− ln(z)− iπ − iπ)

=i

2π(ln(z)− ln(−z)− iπ)

=1

2+ i


.

And thus the explicit form of the Dirac Delta function is

δ(z) =i

2π

(1

z− 1

z

).

We generalize the explicit form of the Dirac Delta function taking an argument of a function g(z),provided g(z) is continuously differentiable with roots in the set of real numbers. Often g(z) issimply of the form z − c with c a point in the set of real numbers. Any further form of g(z) can bereduced down to a series of Dirac Delta functions with argument z − c, as given by property (A.3)in Appendix A. For such an argument, the Dirac Delta function appears as

(7.4) δ(g(z)) =i

2π

(1

g(z)− 1

g(z)

).

The presence of 1/z − 1/z is an odd one at first, but allows a segue to our discussion of irreducibleforms. Furthermore, compare the similarities between the derived explicit form of δ(z) to theassembly of its hyperfunction representation as given in [20], [9] and [10]. Additionally in AppendixB, we prove the explicit form (7.3) holds all the properties of the Dirac Delta function tabulated inAppendix A. �

8. Irreducible Forms

To continue our discussion upon the Dirac Delta function with clarity (that is, to explain thepresence of 1/z − 1/z) we must first examine irreducible forms. Irreducible forms preserve indeter-minacies arising in the range of a function that is typically reduced to a simpler function (and assuch, the indeterminacy lost). Consider z/z which is quite often reduced to z/z = 1; however z/zevaluated at the point z = 0 yields the indeterminacy 0/0, though 1 evaluated at 0 is simply 1.


By defining irreducible forms when the range of a function contains an indeterminacy we are ableto retain sound mathematical consistency. This is exemplified in Sections 10.2 and 10.3, where inSection 10.3 we are able to directly derive 4πδ3(~r) as the Laplacian of 1/r by using an irreducibleform. We consider the irreducible forms of two functions

f1(z) =z

z,

and

f2(z) =1

z− 1

z.

We have already encountered f2(z) in the derivation of the Dirac Delta function and f1(z) appearsin the Laplacian of 1/r in Section 10.3. It is easy to see these functions have irreducible forms asthey can be reduced to 1, and 0 respectively, however at z = 0 both return the indeterminate forms0/0 for f1(z) and ∞−∞ for f2(z). We begin with f1(z).

Lemma 3. Let z ∈ R, then the irreducible form of f1(z) appears as

(8.1)z

z= θ(z) + θ(−z).

Proof. We begin with the following relation

sgn(|z|) =|z|||z||

=z

z.

We consider (4.3), which gives sgn(|z|) = θ(|z|) − θ(−|z|). By interpreting θ(−|z|), we are able toreduce the Signum function with an absolute value argument to sgn(|z|) = θ(|z|). The image of |z|is the nonnegative real numbers, thus θ(−|z|) is equal to 0 for all z ∈ R except z = 0 at which ittakes an indeterminate value. At z = 0, θ(z) equally takes an indeterminate value, therefore wemay remove −θ(−|z|) without losing the indeterminacy. As such, we are left with sgn(|z|) = θ(|z|).Since the absolute value in the Heaviside Step function deters it from ever evaluating a negativenumber, θ(|z|) never equals zero for all possible values z. In fact, θ(|z|) is equal to 1 everywhereexcept at z = 0. We can arrive at an equal image by adding the Heaviside Step function, θ(z), toitself reflected about the ordinate, θ(−z), giving us 1 for z > 0 and z < 0, and an indeterminatevalue at z = 0. Restated,

sgn(|z|) = θ(|z|)= θ(z) + θ(−z).

Thus,

sgn(|z|) =|z|||z||

|z|||z||

= θ(z) + θ(−z).

We conclude

(8.2)z

z= θ(z) + θ(−z).

�

If we replace z/z with unity in the above equation, this mirrors the identity θ(z) = 1 − θ(−z).Furthermore, we make note that zn/zn for all n has an equivalent irreducible form as z/z as long

10 M. KYLE MURPHY

as the indeterminacy at z = 0 is kept. We now examine the irreducible form of f1(z) subject todifferential and indefinite integral operators. Differentiation gives us

d

dz

(zz

)=

d

dz(θ(z) + θ(−z))(8.3)

= δ(z)− δ(z).We cannot subtract out δ(z) without losing the indeterminate value at z = 0. We must now attendto a very important corollary involving the Dirac Delta function.

Corollary 2. δ(z) = −δ(z).

Proof.

i

2π

(1

z− 1

z

)=−i2π

(1

z− 1

z

)=

i

2π

(−1

z+

1

z

)=

i

2π

(1

z− 1

z

)δ(z) = −δ(z).

Thus, since i−1 = −i, the imaginary unit is mutable between the numerator and denominator

i

2π

(1

z− 1

z

)=

1

2πi

(1

z− 1

z

);

(8.4) δ(z) = −δ(z).�

Remark. This is actually to be expected. The Dirac Delta function is 0 everywhere except for asingle point at which it takes an indeterminate value. Consider as a supplementary explanation−0 = +0 = 0 the neutrality of zero. Further, consider the function at z = 0, when the functiontakes the indeterminate value∞−∞ = −∞+∞. Just like 0, the Dirac Delta function has no sign.

Continuing with (8.3) and identity (8.4)

d

dz

(zz

)= 2δ(z).(8.5)

Integration of the irreducible form f1(z) grants the same result as integration of the reducible form,∫zdz

z=

∫(θ(z) + θ(−z))dz(8.6)

= z + C,with C a constant of integration. We now move to the second irreducible form.

Lemma 4. Let z ∈ R, then the function f2(z) takes the irreducible form

(8.7)1

z− 1

z= 2πiδ(z).

Proof. By (7.3), through simple algebra it is clear

1

z− 1

z= −2πiδ(z).

Using identity (8.4),

1

z− 1

z= 2πiδ(z).


We conclude our treatment of f2(z) by evaluating its derivative and indefinite integral. Differenti-ating the irreducible form

d

dz

(1

z− 1

z

)=

d

dz(2πiδ(z))(8.8)

= 2πiδ′(z).

And integration of f2(z) gives ∫ (1

z− 1

z

)dz =

∫2πiδ(z)dz(8.9)

= 2πi(θ(z) + C).�

In the following section a more rigorous and satisfying assay of the integration of f2(z) is conductedvia borrowed tools from complex analysis. This later analysis highlights the residue-driven selectorproperty of the explicit Dirac Delta function along with a discourse upon the poles present in δ(z).Summarizing the range-specific irreducible forms

(8.10) f1(z) =z

z=

{1 if z ∈ R \ {0}θ(z) + θ(−z) if z ∈ R

,

(8.11) f2(z) =1

z− 1

z=

{0 if z ∈ R \ {0}2πiδ(z) if z ∈ R

.

9. Selector Property of the Explicit Form of the Dirac Delta Function

Definition 6. The selector property arises during integration of the Dirac Delta function; it is thefundamental and defining property of the Dirac Delta function [13]. The tall, thin spike conceptof the Dirac Delta function has often been utilized to explain the selector property, however in thefollowing theorem, we demonstrate it is instead the residue about a set of poles that results in theselector property of (7.3). Stated mathematically, the selector property appears as

(9.1)

∫ b

a

f(z)δ(z − c)dz =

0 if c > b or c < a

f(c) if a < c < b

Indeterminate if c = a or c = b

.

As seen in the piecewise above, provided the roots of the argument of the Dirac Delta functionexist within the bounds of integration, the selector property samples the function f(z) at the rootc. Occasionally we find the Dirac Delta function evaluating arguments other than z − c. In thissection, we denote these functions g(z).

Theorem 9.1. Let g(z) be a continuously differentiable function with non-degenerate roots zi ∈ Rand f(z) any function provided it is continuous at f(zi). For the general selector property of theDirac Delta function

(9.2)

∫ ∞−∞

f(z)δ(g(z))dz =∑i

Resz=zi

(f(z)

|g′(zi)|(z − zi)

)=∑i

f(zi)

|g′(zi)|,

where the summation is performed over the roots of g(z).

The roots of g(z) must be in the set of real numbers. This restriction arises from the integralexisting on the extended real number line. For instance, a g(z) = z2 + 1 which has imaginary rootszi = ±i would yield an integral evaluating to 0, even with bounds −∞,∞ since the selecting propertyof δ(z) would then lie outside of the set of real numbers. Recall our discussion of the point z = 0 inthe functions θ(z), R(z), sgn(z), and |z| which we concluded to be two overlapping, non-removable

12 M. KYLE MURPHY

logarithmic branch points. Since dθ(z)/dz = δ(z), it is anticipated that we have two overlapping,non-removable simple poles at z = 0. We take caution before canceling the poles or branding thepoint z = 0 a multipole; these two overlapping poles do not exhibit normal behavior typical of poles.Take in consideration the limits

limz→0

(1

z− 1

z

)= 0 and lim

z→0δ(z) = 0.

Neither limits explode to infinity as characteristic of a pole; nor is the point z = 0 a removablesingularity as we find its Laurent expansion possesses an existing negative-index coefficient a−1.Conversely, if it was a removable singularity, removing it would eliminate the uniqueness of δ(z)(zero everywhere except at a single point) by replacing it with zero. Additionally, it is not anessential singularity as both limits exist, and its Laurent expansion does not have infinitely-existingnegative-index coefficients. We refine our statement by concluding the point z = 0 is a set ofoverlapping simple poles with the curious property that they do not explode the function underevaluation of the limit. Such a statement supports our findings that θ(z), R(z), sgn(z), and |z| haveoverlapping logarithmic branch points.

Proof. We consider the definite integral∫ ∞−∞

δ(z)dz =

∫ ∞−∞

i

2π

(1

z− 1

z

)dz.(9.3)

We integrate (9.3) via Cauchy’s Residue Theorem∮C

f(z)dz = 2πi∑i

Res(f(z), zi),

with zi the isolated singularities of f(z) contained within the contour of integration. The residueof a function about an isolated singularity is equivalent to the a−1 term of the function’s Laurentseries expansion about the isolated singularity. We use this definition to compute∮

C

i

2π

(1

z− 1

z

)dz,

integrating about the curve C, a simple, counterclockwise closed curve around the singularity z = 0.We now evaluate the residue of 1/z − 1/z by constructing its Laurent series about z = 0. This ismade easy by looking at the Laurent series of 1/z and −1/z. The Laurent series of a function, f(z),about a point c is given with

f(z) =

∞∑n=−∞

an(z − c)n.

For 1/z and −1/z with c = 0

1

z= · · ·+ a−2

z2+

a−1z

+ a0 + a1z + a2z2 + · · ·

−1

z= · · · − a−2

z2− a−1

z− a0 − a1z − a2z2 − · · · .

Since 1/z = 1/z, all the coefficients of both Laurent series are equal; (in fact, all coefficients excepta−1 reckon to 0 for the Laurent series expansion of 1/z. 1/z is its own Laurent series expansion.)adding the two Laurent series we obtain

1

z− 1

z= · · ·+ a−2(z−2 − z−2) + a−1(z−1 − z−1) + 0 + a1(z − z) + · · · .

Removing all coefficients that equate to zero leaves us with

1

z− 1

z= a−1

(1

z− 1

z

),(9.4)


and a−1 = 1. We can also show that a−1 = 1 for 1/z − 1/z by using the irreducible form (8.7) withthe formula for the coefficients of the Laurent expansion

an =1

2πi

∮C

f(z)dz

(z − c)n+1,

provided our curve, C, encloses the singularity z = 0. For n = −1 we compute

a−1 =1

2πi

∮C

2πiδ(z)dz

=

∮C

δ(z)dz

a−1 = 1.

From the equality of coefficients in the Laurent series of 1/z and −1/z and the combination thereofto produce the Laurent series of 1/z − 1/z, along with the integral calculation of a−1, we state

Resz=0

(1

z− 1

z

)= Resz=0

(1

z

).(9.5)

Before we return to the integral, recall the interchangeability of the imaginary unit between numer-ator and denominator permitted by identity (8.4). Thus∮

C

1

2πi

(1

z− 1

z

)dz = Resz=0

(1

z

).(9.6)

Similarly, for the shifted Delta function δ(z − c)

(9.7)

∮C

1

2πi

(1

z − α− 1

z − α

)dz = Resz=α

(1

z − α

).

The Dirac Delta function equates to zero everywhere except z = 0 (and z = c for the shifted form)therefore concerning (9.6) we conclude∮

C

i

2π

(1

z− 1

z

)dz =

∫ ∞−∞

i

2π

(1

z− 1

z

)dz

= Resz=0

(1

z

)= 1.

Secondly, we can prove our result using the irreducible form of 1/z − 1/z∫ ∞−∞

i

2π

(1

z− 1

z

)dz =

1

2πi

∫ ∞−∞

2πiδ(z)dz

= 1.

We generalize for a function, f(z), multiplying δ(g(z)) by starting with∫ ∞−∞

f(z)δ(g(z))dz =

∫ ∞−∞

f(z)∑i

δ(z − zi)|g′(zi)|

dz.

14 M. KYLE MURPHY

from identity (A.3) in Appendix A and given in [13]. We write δ(z) in its explicit form and movethe summation over the roots outside the integral to give∫ ∞

−∞f(z)δ(g(z))dz =

1

2πi

∑i

∫ ∞−∞

f(z)

|g′(zi)|

(1

z − zi− 1

z − zi

)dz

=∑i

Resz=zi

(f(z)

|g′(zi)|(z − zi)

)=∑i

f(zi)

|g′(zi)|.

As such, we have completed the theorem,∫ ∞−∞

f(z)δ(g(z))dz =∑i

Resz=zi

(f(z)

|g′(zi)|(z − zi)

)=∑i

f(zi)

|g′(zi)|.

�

9.1. An Additional Identity of the Dirac Delta Function. In addition to identity (8.4), wederive another identity of the Dirac Delta function.

Lemma 5. The relation between the n-th derivative of the Dirac Delta and the Dirac Delta is

(9.8) δn(z) = n!δ(zn+1).

Proof. Differentiating the explicit form returns

(9.9) δ′(z) =i

2π

(1

z2− 1

z2

).

The negative signs that appear due to the differentiation of 1/z are absorbed via identity (8.4).Further differentiating the Dirac Delta prime explicit form

(9.10) δ′′(z) =2i

2π

(1

z3− 1

z3

),

once further,

(9.11) δ′′′(z) =6i

2π

(1

z4− 1

z4

).

For the n-th derivative of the explicit form

δn(z) =n!i

2π

(1

zn+1− 1

zn+1

)δn(z) = n!δ(zn+1).

�

10. Applications of Irreducible Forms

We investigate the efficacy of irreducible forms by considering two applications which benefit fromthe use of irreducible forms. First, we evaluate the vorticity of a 2-D irrotational vortex with a resultproportional to the 2-D Dirac Delta function. This result is only directly obtainable through the useof an irreducible form. Second, we analyze the famous problem of the Laplacian of 1/r and directlyarrive at the result ∇2(1/r) = −4πδ3(~r) via the inclusion of an irreducible form. Before we begin,we must define the multidimensional Dirac Delta function.


10.1. Multidimensional Dirac Delta Function.

Definition 7. The Dirac Delta function can be defined in higher dimensions and for differentcoordinate systems as laid out in [13][5]. We denote δ3(x, y, z) as the 3-D Dirac Delta function inCartesian coordinates, given by [5] as the relation between products of 1-D Dirac Delta functions

δ3(~r) = δ3(x, y, z) = δ(x)δ(y)δ(z) =

{0 if x2 + y2 + z2 6= 0

Indeterminate if x2 + y2 + z2 = 0.

By equation (7.3), the Dirac Delta function must always evaluate a scalar argument else we wouldhave division by a vector. However to retain consistent notation for the multidimensional DiracDelta function, we denote δn(~r) as the product of Dirac Delta functions of the n components of ~r.For the planar case in polar coordinates

δ2(~r) = δ2(r, θ) = δ(r)δ(θ),

which can be extended to cylindrical coordinates

δ3(~r) = δ3(r, θ, z) = δ(r)δ(θ)δ(z).

And in spherical coordinates

δ3(~r) = δ3(r, θ, φ) = δ(r)δ(θ)δ(φ).

Bracewell gives us relations between the multidimensional Dirac Delta function in polar, cylindrical,and spherical coordinates and the 1-D Dirac Delta function with distance argument δ(r). From [5],between the one dimensional Dirac Delta function and the planar, polar Dirac Delta function withθ-symmetry

δ2(~r) =δ(r)

πr.(10.1)

Multiplying this relation by δ(z) produces the 3-D Dirac Delta function for cylindrical coordinatesand a θ-symmetrical case

(10.2) δ3(~r) =δ(r)

πrδ(z).

Furthermore, Bracewell [5] gives us a connection between δ3(~r) and δ(r) in spherical coordinates forθ-symmetry and φ-symmetry

(10.3) δ3(~r) =δ(r)

2πr2.

10.2. Application: Vorticity and Circulation of a Vortex Flow Field. We examine thevorticity of a planar, irrotational vortex in inviscid flow. Since the vortex is irrotational, we expectthe vorticity to equate to zero everywhere except the center. For a flow velocity vector ~v, vorticity~ω is defined by Anderson in [2] as the following

~ω ≡ ~∇× ~v.

In terms of the circulation, Γ, the flow velocity vector for an irrotational vortex in inviscid flow isgiven by [2] as

~v =−Γ

2πrθ = vθ,

The formula governing the circulation, from [2], is

Γ =

∮∂S

~v · d~l =

∫ ∫S

~ω · d~S.

16 M. KYLE MURPHY

For the curl of the velocity vector in cylindrical coordinates

~∇× ~v =1

r

∂

∂r(rvθ)z

=1

r

∂

∂r

(−Γr

2πr

)z

=−Γ

2πr

∂

∂r

(rr

)z.

We replace the function r/r with its irreducible form (8.2) and evaluate the derivative from (8.5),

~∇× ~v =−Γ

2πr

∂

∂r(θ(r) + θ(−r))z

=−Γδ(r)

πrz.

We absorb the negative sign with identity (8.4): δ(r) = −δ(r), and write in terms of the 2-D polarDirac Delta function from (10.1)

~∇× ~v = Γδ2(~r)z.

We check our solution to the vorticity by deriving the circulation Γ of the vortex. From the definitionof circulation,

Γ =

∫ ∫S

~ω · d~S

=

∫ 2π

0

∫ R

0

Γδ(r)

πrrdrdθ

= 2

∫ R

0

Γδ(r)dr.

We alter the bounds of integration to include 0 rather than 0 at one of the bounds. The Dirac Deltafunction is an even function which legalizes the change, giving us

Γ = Γ

R∫−R

δ(r)dr

= Γ.

And thus the vorticity of an irrotational vortex in inviscid flow can be defined as

(10.4) ~ω =Γδ(r)

πrz = Γδ2(~r)z.

10.3. Application: Laplacian of 1/r. To further strengthen the usefulness of irreducible formswe evaluate the Laplacian of 1/r. Let t = 1/r so that

∇2t = ∇2 1

r.

Operating in spherical coordinates, and for our example

∇2t =1

r2∂

∂r

(r2∂t

∂r

),

with∂t

∂r= − 1

r2.

The equation for the Laplacian in spherical coordinates becomes

∇2t =1

r2∂

∂r

(−r2

r2

).


We reduce −r2/r2 to −r/r without losing the indeterminacy at 0. By then replacing r/r with itsirreducible form (8.2) and differentiating

∇2t =1

r2∂

∂r(−θ(r)− θ(−r))

=1

r2(−δ(r)− (−δ(r)).

And from identity (8.4)

∇2t =−2δ(r)

r2.(10.5)

We seek to replace the 1-D Dirac Delta function in (10.5) with the spherical, 3-D Dirac Delta functionby use of relation (10.3). Solving for δ(r) in terms of δ3(~r) from (10.3) returns

δ(r) = 2πr2δ3(~r).

Substituting this result into (10.5) completes our example

∇2t =−2δ(r)

r2

=−2(2πr2δ3(~r))

r2

= −4πδ3(~r).

Thus we have directly confirmed the Laplacian of 1/r without having to assemble the answer toagree with the divergence theorem as discussed in [8]. Thus

(10.6) ∇2t = ∇2 1

r= −4πδ3(~r)

in concordance with [8]. Due to identity (8.4), this may also be written as

∇2t = ∇2 1

r= 4πδ3(~r).

11. Acknowledgments

I would like to thank Professor Itzhak Bars of the University of Southern California and ProfessorDavid J. Brown of North Carolina State University for the series of classes and lectures that developedmy skills needed for this paper. I must also acknowledge the University of Southern California forgiving me the ideal forum in which to compose this paper. Importantly, I thank Taina M. Colon forher support and friendship.

12. Appendix

12.1. Appendix A: Properties of the Dirac Delta Function. The Dirac Delta holds manyproperties, some of which concern the Delta function by itself, its derivatives, and its appearance inan integrand. Below are listed some of the fundamental and prominent properties of δ(z) as given

18 M. KYLE MURPHY

in [5][13][19][7]

δ(z) =dθ(z)

dz(A.1)

δ(z − α) =

{0 if z 6= α

Indeterminate if z = α(A.2)

δ(g(z)) =∑i


=

{0 if z 6= zi

Indeterminate if z = zi(A.3)

δ(αz) =1

|α|δ(z)(A.4)

δ(z2 − α2) =1

2|α|(δ(z − α) + δ(z + α))(A.5)

δ(−z) = δ(z)(A.6)

δ′(z)z = −δ(z)(A.7)

δ′(−z) = −δ′(z)(A.8)

dnδ(z)

dzn(zn) = n!(−1)nδ(z)(A.9) ∫ ∞

−∞δ(z)dz =

∫ ∞−∞

δ(z − α)dz = 1(A.10) ∫ b

a

δ(z − α)dz =

{0 if α < a, α > b

1 if a < α < b(A.11) ∫ ∞

−∞f(z)δ(z − α)dz = f(α)(A.12) ∫ b

a

f(z)δ(z − α)dz =

{0 if a > α, b < α

f(α) if a < α < b(A.13) ∫ ∞

−∞f(z)δ′(z)dz = −f ′(0)(A.14) ∫ ∞

−∞f(z)δn(z)dz = −

∫ ∞−∞

∂f(z)

∂zδn−1(z)dz = (−1)nfn(0)(A.15)

We note g(z) is a continuously differentiable function with non-degenerate roots zi.

12.2. Appendix B: Proofs of the Congruency of Dirac Delta Properties. We have alreadyestablished the property (A.1) in Section 7 as it was paramount in the proof of the explicit form(7.3).

A.2.

δ(z − α) =i

2π

(1

z − α− 1

z − α

)=

{0 if z 6= α

Indeterminate if z = α

�

A.3.

δ(g(z)) =i

2π

(1

g(z)− 1

g(z)

)=

{0 when z 6= zi

Indeterminate when z = zi,


where g(z) is a continuously differentiable function with non-degenerate roots zi. Equally,


=∑i

i

2π|g′(zi)|

(1

z − zi− 1

z − zi

)=

{0 when z 6= zi

Indeterminate when z = zi,

where the summation is performed over the roots zi. Recall the conclusion of (9.5) that the residueof 1/z − 1/z is the residue of 1/z. Since g(z) must have non-degenerate roots, we conclude 1/g(z)to be composed of simple poles, thus for the residue∑

i

Resz=zi

(1

g(z)

)=∑i

limz→zi

z − zig(z)

=∑i

1

g′(zi).

Due to identity (8.4) we take the absolute value of g′(zi), obtaining∑i

Resz=zi

(1

g(z)

)=∑i

1

|g′(zi)|.

Additionally, for the decomposed form of δ(g(z))∑i

Resz=zi

(1

|g′(zi)|(z − zi)

)=∑i

limz→zi

z − zi|g′(zi)|(z − zi)

=∑i

1

|g′(zi)|.

And we have shown the explicit form holds for (A.3). �

A.4.

δ(αz) =i

2π

(1

αz− 1

αz

).

The sign of z does not change our function, as evident by (8.4) thus we can strip the sign off α anddeposit it upon z

i

2π|α|

(1

±z− 1

±z

)=

1

|α|i

2π

(1

z− 1

z

)δ(αz) =

1

|α|δ(z).

�

A.5.

δ(z2 − α2) =i

2π

(1

z2 − α2− 1

z2 − α2

).

We perform the partial fraction decomposition

1

z2 − α2=

1

(z − α)(z + α)=

A

z − α+

B

z + α,

and solve for the coefficients A,B giving A = 1/2α and B = −1/2α. Substituting in the values ofA,B admits

1

z2 − α2=

1

2α(z − α)− 1

2α(z + α).

20 M. KYLE MURPHY

We return to the expression for δ(z2 − α2) using the partial fraction decomposition

δ(z2 − α2) =i

2π

[(1

2α(z − α)− 1

2α(z + α)

)−(

1

2α(z − α)− 1

2α(z + α)

)]=

1

2|α|i

2π

(1

z − α− 1

z − α+

1

z + α− 1

z + α

)δ(z2 − α2) =

1

2|α|(δ(z − α) + δ(z + α)).

�

Identity (A.6) for the explicit form follows from identity (8.4).

A.7. For the explicit form of δ′(z) we reference (9.9). Multiplying by z and using identity (8.4)

i

2π

( zz2− z

z2

)=

i

2π

(1

z− 1

z

)δ′(z)z = −δ(z).

�

A.8. From (9.9),

δ′(−z) =i

2π

(1

(−z)2− 1

(−z)2

)=

i

2π

(1

z2− 1

z2

)= δ′(z).

Identity (8.4) holds for all derivatives of the Dirac Delta, thus

δ′(−z) = −δ′(z)

�

A.9. Multiplying the n-th derivative of the Dirac Delta function by zn, and using identity (9.8),

n!δ(zn+1)zn =n!i

2π

(zn

zn+1− zn

zn+1

).

We reduce the fractions to 1/z and absorb the (−1)n term in (A.9) by identity (8.4) to render

n!δ(zn+1)zn =n!i

2π

(1

z− 1

z

)= n!δ(z).

�

Identity (A.10) for the explicit form is proven in Section 9 in equations (9.6) and (9.7). For (A.11)the bounds of integration must encompass the poles in order to obtain the residue about the poles,else the integral evaluates to zero.


A.12. ∫ ∞−∞

f(z)δ(z − α)dz =1

2πi

∫ ∞−∞

(f(z)

z − α− f(z)

z − α

)dz

= Resz=αf(z)

z − α= f(α).∫ ∞

−∞f(z)δ(z − α)dz = f(α).

�

The argument concerning the proof of identity (A.13) for the explicit form follows the argumentof (A.11): the bounds of integration must contain the poles of the Dirac Delta function in order totake their residues.

A.14. ∫ ∞−∞

f(z)δ′(z)dz = −f ′(0).

Expanding δ′(z) to its explicit form (9.9) transforms the integral to

1

2πi

∫ ∞−∞

f(z)

(1

z2− 1

z2

)dz.

Shifting a factor of 1/z outside of 1/z2 − 1/z2 (echoing identity (8) for n = 1) gives

1

2πi

∫ ∞−∞

f(z)

z

(1

z− 1

z

)dz = Resz=0

(f(z)

z2

).

For the residue at a multipole of order 2

Resz=0

(f(z)

z2

)= limz→0

d

dz

z2f(z)

z2.

Since the limit approaches 0 the range does not contain the point of indeterminacy; it is unnecessaryto use the irreducible form of z/z when evaluating the residue. Reducing z2/z2 to 1 gives

Resz=0

(f(z)

z2

)= f ′(0).

By identity (8.4), ∫ ∞−∞

f(z)δ′(z)dz = −f ′(0).

�

A.15. We have shown in (9.8) that δn(z) = n!δ(zn+1). From this we rewrite the integrand of (A.15)to ∫ ∞

−∞f(z)n!δ(zn+1)dz =

n!i

2π

∫ ∞−∞

f(z)

(1

zn+1− 1

zn+1

)dz.

22 M. KYLE MURPHY

The integral may be evaluated by the residue theorem, where we take the residue about the pole oforder n+ 1. Thus

n!i

2π

∫ ∞−∞

f(z)

(1

zn+1− 1

zn+1

)dz =

n!i

2π· 2πi · Resz=0

(f(z)

zn+1

)= limz→0

dn

dzn(zn+1)

f(z)

zn+1∫ ∞−∞

f(z)n!δ(zn+1)dz = fn(0).

�

Notice we have absorbed the (−1)n term due to identity (8.4).

12.3. Appendix C: Fourier Transformations. We compute the frequency Fourier transforma-tions of the explicit forms of θ(z), R(z), sgn(z), |z|, and δ(z) to show that they agree with theaccepted Fourier transforms of these functions tabulated in [4][18]. We then end on the Fouriertransforms of the two irreducible forms presented. The forward frequency Fourier transform of afunction f(x) is defined as

Fx[f(x)](k) = F (k) =

∫ ∞−∞

f(x)e−2πikxdx.

Listed below are the Fourier transforms of the functions which we wish to prove:

Fz[θ(z)](k) =δ(k)

2− i

2πk(C.1)

Fz[R(z)](k) =i

4πδ′(k)− 1

4π2k2(C.2)

Fz[sgn(z)](k) =−iπk

(C.3)

Fz[|z|](k) =−1

2π2k2(C.4)

Fz[δ(z − α)](k) = exp(−2πikα)(C.5)

Heaviside Step Function. We do not expect the Fourier transform of θ(z) to converge; it is typicallyassigned its transform from the Fourier transform of a limit of an approximating function as demon-strated in [6]. Through our own ansatz, we can arrive at the desired transform F (k). Consideringthe explicit form of the Heaviside Step function

Fz[θ(z)](k) = Fz[

1

2+ i


](k)

=

∫ ∞−∞

(1

2+ i


)e−2πikzdz.

Since all z < 0 gives θ(z) = 0, we may alter the bounds of integration from −∞,∞ to ε,∞ where εapproaches 0, since θ(0) is indeterminate. Then

F (k) = limε→0

∫ ∞ε

(1

2+ i


)e−2πikzdz

= limε→0

∫ ∞ε

e−2πikz

2dz + lim

ε→0

i

2π

∫ ∞ε

(ln(z)− ln(−z))e−2πikzdz,


and since z > 0

Fz[

1

2

](k) + lim

ε→0

i

2π

∫ ∞ε

(ln(z)− ln(z)− iπ)e−2πikzdz

=δ(k)

2+ limε→0

−i2π2π

∫ ∞ε

e−2πikzdz

=δ(k)

2+

i

4πklimε→0

[e−2πikz]∞ε =δ(k)

2+

i

4πk

[limz→∞

e−2πikz − 1].

The limit limz→∞ e−2πikz has an oscillation of 2 at∞ and thus the limit does not exist, contributingto the expected non-convergence of the Fourier transform of θ(z). As z → ∞ e−2πiz oscillates

between e−2πi0 = 1, e−2πi12 = −1, and e−2πi1 = 1, thus we can say z oscillates between 0 and 1.

Our ansatz is to evaluate e−2πiz at the average of oscillation z = 1/2, yielding e−πi = −1 and thus

δ(k)

2+

i

4πk

[e−πi − 1

]=δ(k)

2+

i

4πk[−1− 1],

F (k) =δ(k)

2− i

2πk,

which agrees with (C.1). �

Ramp Function. We have demonstrated the Fourier transform of our explicit form of θ(z) (grantedwith limited assumptions) is the Fourier transform of θ(z) (which is equally determined with limitedassumptions). Due to the properties of the Fourier transformation, it is unnecessary to computethe complicated integrals arising for each one of our remaining functions. We now seek the Fouriertransform of R(z), recalling (3.2): R(z) = zθ(z),

Fz[R(z)](k) = Fz[zθ(z)](k)

= Fz[z

(1

2+ i


)](k).

Utilizing the Fourier transform property

Fz[znf(z)](k) =

(i

2π

)ndnF (k)

dkn

in evaluation of the Ramp function gives

Fz[zθ(z)](k) =i

2π

dΘ(k)

dk.

We define Θ(k) as the Fourier transform of θ(z). Performing our transformation results

F (k) =i

2π

d

dk

(δ(k)

2− i

2πk

)=

i

2π

(δ′(k)

2+

i

2πk2

)F (k) =

iδ′(k)

4π− 1

4π2k2.

�

24 M. KYLE MURPHY

Signum Function. The linearity property of the Fourier transform enables a fairly easy and quickcomputation of F (k) for sgn(z). We know from (4.2) sgn(z) = 2θ(z)− 1 so

Fz[sgn(z)](k) = Fz[2θ(z)− 1](k)

= 2Fz[θ(z)](k)−Fz[1](k)

= 2

(δ(k)

2− i

2πk

)− δ(k),

F (k) =−iπk.

�

Absolute Value Function. The method of computing F (k) for |z| is conducted equally as F (k) forR(z). From (5.2) |z| = zsgn(z) so that

Fz[|z|](k) = Fz[zsgn(z)](k)

=i

2π

dSGN(k)

dk.

We note SGN(k) as the Fourier transform of sgn(z) giving

i

2π

d

dk

(−iπk

)=

i

2π

i

πk2

F (k) =−1

2π2k2.

�

Dirac Delta Function. We consider the shifted Dirac Delta function and take F (k) of δ(z − α)

Fz[δ(z − α)](k) = Fz[i

2π

(1

z − α− 1

z − α

)](k)

=

∫ ∞−∞

i

2π

(1

z − α− 1

z − α

)e−2πikzdz.

We can evaluate this integral via the Residue Theorem, recalling (9.5) for the shifted Dirac Delta

Resz=α

(1

z − α− 1

z − α

)= Resz=α

1

z − α= 1,

and so

1

2πi

∫ ∞−∞

(1

z − α− 1

z − α

)e−2πikzdz

= Resz=α

(exp(−2πikz)

z − α

),

F (k) = e−2πikα.

�

Irreducible Forms. Moving to the Fourier transforms of the irreducible forms of f1(z) and f2(z), weevaluate

Fz[zz

](k) = Fz[θ(z) + θ(−z)](k)

=δ(k)

2− i

2πk+ Fz[θ(−z)](k)

=δ(k)

2− i

2πk+δ(k)

2+

i

2π

∫ ∞−∞

(ln(−z)− ln(z))e−2πikzdz.


For z > 0, θ(−z) always returns 0, therefore we may alter the bounds of integration as before

δ(k)

2− i

2πk+δ(k)

2+

i

2πlimε→0

∫ −ε−∞

(ln(z) + iπ − ln(z))e−2πikzdz

=δ(k)

2− i

2πk+δ(k)

2+i2π

2πlimε→0

∫ −ε−∞

e−2πikzdz

=δ(k)

2− i

2πk+δ(k)

2− i

4πk[−1− e−2πikz]z=−∞.

We make the same ansatz as previous when evaluating Fz[θ(z)](k), that is: z = 1/2 for exp(−2πiz).Plugging in returns

δ(k)

2− i

2πk+δ(k)

2− i

4πk[−1− 1]

=δ(k)

2− i

2πk+δ(k)

2+

i

2πk= δ(k).

And therefore the Fourier transform of the irreducible form of z/z is identical to the Fourier transformof its reducible form, 1. Resolving Fz[f2(z)](k)

Fz[

1

z− 1

z

](k) =

∫ ∞−∞

(1

z− 1

z

)e−2πikzdz

= 2πi · Resz=0

(e−2πikz

z

)= 2πi(e2πik0),

Fz[

1

z− 1

z

](k) = 2πi.

For the shifted f2(z − c)

Fz[

1

z − c− 1

z − c

](k) =

∫ ∞−∞

(1

z − c− 1

z − c

)e−2πikzdz

= 2πi · Resz=c

(e−2πikz

z − c

)Fz[

1

z − c− 1

z − c

](k) = 2πi · e−2πikc.

�

References

[1] Abramowitz, Milton and Irene Stegun, eds. Handbook of Mathematical Functions with Formulas, Graphs, and

Mathematical Tables. Washington, D.C.: National Bureau Of Standards, 1972. Print.[2] Anderson, Jr., John D. ”Fundamentals of Inviscid, Incompressible Flow.” Ch. 3 in Fundamentals of Aerodynamics,

5th ed. New York: McGraw-Hill, 2011. 262-265. Print.

[3] Balakrishnan, V.. ”All about the Dirac Delta Function(?).” 1-10. Web. Feb. 2015.¡http://www.physics.iitm.ac.in/ labs/dynamical/pedagogy/vb/delta.pdf¿.

[4] Bevel, P J. ”Fourier Transform Pairs.” The Fourier Transform. N.p., n.d. Web. Mar. 2015.

¡http://www.thefouriertransform.com/pairs/fourier.php¿.[5] Bracewell, R. ”The Impulse Symbol.” Ch. 5 in The Fourier Transform and Its Applications, 3rd ed. New York:

McGraw-Hill, pp. 74-104, 1999.[6] Cuff, Paul. ”Powerpoint.” Lecture 8. Princeton. Web. Mar. 2015.

¡http://www.princeton.edu/ cuff/ele301/files/lecture8 2.pdf¿.

[7] Dirac, P. A. M.. The Principles of Quantum Mechanics. London: Oxford University Press, 1958. Print.[8] Griffiths, David J. ”The Dirac Delta Function.” Introduction to Electrodynamics, 3rd Ed. New Jersey: Prentice-

Hall, 1999. 45-50. Print.

26 M. KYLE MURPHY

[9] Jacobs, Bryan. ”Hyperfunction.” From MathWorld–A Wolfram Web Resource, created by Eric W. Weisstein.http://mathworld.wolfram.com/Hyperfunction.html

[10] Hyperfunction. A. Kaneko (originator), Encyclopedia of Mathematics. URL:

http://www.encyclopediaofmath.org/index.php?title=Hyperfunctionoldid=16339[11] Katz, Mikhail G., and David Tall. ”A Cauchy-Dirac Delta Function.” (2012): 1-24. Print.

[12] Venetis, J.. ”An Analytic Exact Form of the Unit Step Function.” Hrpub. Horizon Research Publishing Corpo-

ration, 2014. Web. Dec. 2014. ¡http://www.hrpub.org/download/20141001/MS2-13402519.pdf¿.[13] Weisstein, Eric W. ”Delta Function.” From MathWorld–A Wolfram Web Resource.

http://mathworld.wolfram.com/DeltaFunction.html[14] Weisstein, Eric W. ”Heaviside Step Function.” From MathWorld–A Wolfram Web Resource.

http://mathworld.wolfram.com/HeavisideStepFunction.html

[15] Weisstein, Eric W. ”Ramp Function.” From MathWorld–A Wolfram Web Resource.http://mathworld.wolfram.com/RampFunction.html

[16] Weisstein, Eric W. ”Absolute Value.” From MathWorld–A Wolfram Web Resource.

http://mathworld.wolfram.com/AbsoluteValue.html[17] Weisstein, Eric W. ”Sign.” From MathWorld–A Wolfram Web Resource.

http://mathworld.wolfram.com/Sign.html

[18] Weisstein, Eric W. ”Fourier Transform.” From MathWorld–A Wolfram Web Resource.http://mathworld.wolfram.com/FourierTransform.html

[19] Wheeler, Nicholas. ”Dirac Delta Function Identities.” 1-18. Web. Feb. 2015.

[20] Zharinov, Victor. ”Hyperfunctions and Analytic Functionals.” 317-333. Web. Mar. 2015.¡http://oai.cwi.nl/oai/asset/18237/18237B.pdf¿.

Department of Astronautical Engineering, USC, Los Angeles California 90089

Email Address: [email protected]

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