Mathematica examples relevant to Gamma and Beta functions

Gamma function: Gamma[x]Check that the defining integral indeed gives Gamma function

Integrate[x^(p -− 1) Exp[-−x], {x, 0, Infinity}, Assumptions → Re[p] > 0]

Gamma[p]

Check recursion relation (following quantity should equal 1)

check[p_] = Gamma[p] p /∕ Gamma[p + 1];

Plot[check[p], {p, -−10, 10}, PlotStyle → {Red, Thick}]

-−10 -−5 5 10

0.0

0.5

1.5

2.0

Gamma[p] is indeed (p-1)! for integer p:

{Gamma[7], 6!}

{720, 720}

Plot shows the poles in the Gamma function on the real axis.

Plot[Gamma[x], {x, -−3, 3}, PlotStyle → Thick]

-−3 -−2 -−1 1 2 3

-−10

-−5

5

10

Here’s a 3D plot of the absolute value of the Gamma function in the complex plane. Note that you can rotate the view around.Note the poles at x=0, -1, -2, -3,...

Here’s a 3D plot of the absolute value of the Gamma function in the complex plane. Note that you can rotate the view around.Note the poles at x=0, -1, -2, -3,...

Plot3D[Abs[Gamma[x + I y]], {x, -−4, 4}, {y, -−4, 4},PlotRange → {-−1, 10}, AxesLabel → {"Re[z]", "Im[z]", "Abs[Gamma[z]]"}]

This is the argument---a rather complicated plot!

Plot3D[Arg[Gamma[x + I y]], {x, -−5, 5}, {y, -−5, 5},PlotRange → {-−π, π}, AxesLabel → {"Re[z]", "Im[z]", "Arg[Gamma[z]]"}]

Notice that Arg[Gamma[z]]=0 when z is real and positive, as expected.

Keep in mind that Arg is between -π𝜋 and π𝜋. We can see the “sheets” are connected up to factors of 2π𝜋.

Plot3D[{Arg[Gamma[x + I y]], Arg[Gamma[x + I y]] + 2 π, Arg[Gamma[x + I y]] + 4 π},{x, -−5, 5}, {y, -−5, 5}, AxesLabel → {"Re[z]", "Im[z]", "Arg[Gamma[z]]"}]

Here follows the real and imaginary parts---a more complicated structure emerges around the poles.

2 RevisedGamma.nb

Plot3D[Re[Gamma[x + I y]], {x, -−4, 4}, {y, -−4, 4},PlotRange → {-−10, 10}, AxesLabel → {"Re[z]", "Im[z]", "Re[Gamma[z]]"}]

Plot3D[Im[Gamma[x + I y]], {x, -−4, 4}, {y, -−4, 4},PlotRange → {-−10, 10}, AxesLabel → {"Re[z]", "Im[z]", "Im[Gamma[z]]"}]

For comparison a single pole

Plot3D[Abs[1 /∕ (x + I y)], {x, -−3, 3}, {y, -−3, 3},PlotRange → {-−1, 5}, AxesLabel → {"Re[z]", "Im[z]", "Abs[1/∕z]"}]

Here’s the real part , which is x/(x^2+y^2)

RevisedGamma.nb 3

Plot3D[Re[1 /∕ (x + I y)], {x, -−3, 3}, {y, -−3, 3},PlotRange → {-−2, 2}, AxesLabel → {"Re[z]", "Im[z]", "Re[1/∕z]"}]

...and the imaginary part which is -y/(x^2+y^2)

Plot3D[Im[1 /∕ (x + I y)], {x, -−3, 3}, {y, -−3, 3},PlotRange → {-−10, 10}, AxesLabel → {"Re[z]", "Im[z]", "Im[1/∕z]"}]

Qualitatively these should make sense. The real part flips sign as x goes from negative to positive and the imaginary part flips sign as y goes from negative to positive.

Beta function: Beta[x,y]The following integral defines Beta[x,y] for Re[p,q]>0Mathematica jumps directly to the expression for Beta in terms of Gamma functions

Integrate[x^(p -− 1) (1 -− x)^(q -− 1), {x, 0, 1}, Assumptions → p > 0 && q > 0]

Gamma[p] Gamma[q]

Gamma[p + q]

Checking relation between Gamma and Beta functions

{Beta[.5, .6], Gamma[.5] Gamma[.6] /∕ Gamma[.5 + .6]}

{2.7745, 2.7745}

You can get the definition of the Beta function out of Mathematica by using FunctionExpand[ ]

FunctionExpand[Beta[p, q]]

Gamma[p] Gamma[q]

Gamma[p + q]

4 RevisedGamma.nb

Lets try visualizing the Beta function

-−1

5

Lets look at slices, holding one variable fixed.

Manipulate[Plot[Beta[p, q], {q, -−10, 10}], {p, 5, -−5}]

p

-−10 -−5 5 10

-−1.0

-−0.5

0.5

1.0

To look at the contours, varying both p and q

ContourPlot[Re[Beta[p, q]], {p, -−5, 5}, {q, -−5, 5}, Contours → 15]

RevisedGamma.nb 5

The contour plot is a little easier to look at compared to the 3D plot

Plot3D[Re[Beta[p, q]], {p, -−5, 5}, {q, -−5, 5}]

11.7 #2

IntegrateSqrt[Sin[x]^3 Cos[x]], x, 0, Pi 2

π

4 2

Matching this to the trig form of the Beta funciton (Boas equation 7.6.2):

B (p, q) = 2 ∫π𝜋/∕2

0sin(θ𝜃)2 p-−1 cos(θ𝜃)2 p-−1 dθ𝜃

2p-1 = 3/2 , 2q-1 = 1/2

1

2Beta[5 /∕ 4, 3 /∕ 4]

1

2Gamma

3

4 Gamma

5

4

% /∕/∕ FunctionExpand

π

4 2

6 RevisedGamma.nb