# The complement formula of the Gamma function

1. Jul 7, 2012

### Millennial

Now, I already know the complement formula and I also know a proof that can be presented using the Euler product forms of the Gamma function. What I am curious about is, can one obtain the formula using the integral forms of the Gamma and Beta functions?

I will show my work so far. I started with the identity that implies that the Beta function is an analytic continuation of the binomial coefficients, that is,
$$B(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$
Next, I plugged in 1-a for b to obtain
$$B(a,1-a)=\Gamma(a)\Gamma(1-a)$$
Note that the right-hand side is exactly what the complement formula tells the value of. So, it remains to obtain a simple expression for the Beta function on the left. This is easily seen to be equal to
$$\Gamma(a)\Gamma(1-a) = 2\int_{0}^{\pi/2}\tan^{2a-1}(\theta)d\theta$$
So my question is, can we evaluate this integral? (that is, without using the product representations.)
I already know the value of the integral from the complement formula, and I verified that the path I trace is correct in the sense that Wolfram gives the exact complement formula when evaluating this integral. I want to know if it is possible to do it by hand.

2. Jul 7, 2012

### chiro

Hey Millennial.

I'm wondering if a conversion to the euler identity might help. Since the cos(x) = Re(e^(ix)) and sin(x) = Im(e^(ix)), you can then get this in terms of a function of exponentials (that include imaginary numbers) raised to a given power.

So we know cos(x) = (e^(ix) + e^(-ix))/2 and sin(x) = (e^(ix) - e^(-ix)/2 which means tan(x) = 1 - 2e^(ix)/[e^(ix)+e^(-ix)]. Now when you raise to that some power, you get a form of (1 - f(x))^n for some appropriate n which has a form of a generalized binomial.

Also, another approach might be to draw from what is known as the Gudermannian:

http://en.wikipedia.org/wiki/Gudermannian_function

At the very least, you should be able to manipulate the above to get a series expansion which means that while it won't be in terms of elementary functions, you will still be able to get something useful in terms of being able to get something within some kind of error bound.

3. Jul 7, 2012

### Mute

Hm, it might be tricky. Integrals of powers of sine or cosine are typically only given as recursive formulas.

Continuing with what chiro was saying, though, you might be able to evaluate it as a contour integral. If you write tan in terms of $\exp(i\theta)$,

$$\int_0^{\pi/2} d\theta \tan^{2a-1}\theta = \int_0^{\pi/2} d\theta \left( \frac{e^{i\theta} - e^{-i\theta}}{e^{i\theta}+e^{-i\theta}}\right)^{2a-1}.$$
Then, make the change of variables $z = \exp(i\theta)$ gives a contour integral

$$\oint_A \frac{dz}{iz} \left(\frac{z - z^{-1}}{z + z^{-1}}\right)^{2a-1} = \oint_A \frac{dz}{iz} \left(\frac{z^2 - 1}{z^2 + 1}\right)^{2a-1},$$
where A is the arc of a unit circle from angles 0 to pi/2. (Actually, it doesn't quite hit pi/2 because there is a pole there at z = i).

As written, we can't do much with this contour integral. To actually evaluate it, we would need to consider an integral over a closed contour. So, we close the contour by going along the real and imaginary axis, with a small semicircle of radius $\epsilon$ around the origin to avoid the pole there, as well as a similar semicircle around z = i to avoid that pole. So, if we call the closed contour $\gamma$, the integral to evaluate is

$$\begin{eqnarray*} \oint_\gamma \frac{dz}{iz} \left(\frac{z - z^{-1}}{z + z^{-1}}\right)^{2a-1} & = & \int_0^{\pi/2} d\theta \tan^{2a-1}\theta + \int_0^1 \frac{dx}{ix} \left(\frac{x^2 - 1}{x^2 + 1}\right)^{2a-1} - \int_1^0 \frac{dy}{y} \left(\frac{y^2 + 1}{y^2 - 1}\right)^{2a-1} \\ & & + \int_{\pi/2}^0 \left[\frac{dz}{iz} \left(\frac{z^2 - 1}{z^2 +1}\right)^{2a-1}\right]_{z = \epsilon \exp(i\theta)} + \int_{2\pi}^{3\pi/2} \left[\frac{dz}{iz} \left(\frac{z^2 - 1}{z^2 + 1}\right)^{2a-1}\right]_{z = i + \epsilon \exp(i\theta)}\\ \end{eqnarray*}$$

The last two terms on the right hand side are the contributions of the two pole-avoiding arcs, where the limit $\epsilon \rightarrow 0$ is to be taken. Because the contour encloses no poles, the left hand side is zero, so you can relate the desired integral to the other ones.

That said, you may not be able to evaluate the other integrals very easily. I haven't verified myself whether or not this approach will work directly. It may just give you the integral you want in terms of other integrals you can't do directly.

(You'll also want to double-check my work here).

4. Jul 7, 2012

### Millennial

I came up with another way after struggling to obtain an alternate form of the Beta function.

In the defining integral for the gamma function, I make the substitution $t=\frac{s}{s+1}$. This yields the two limits of integration as infinity and zero, respectively. The integrand can easily be manipulated to yield an alternate form as follows:
$$B(a,b)=\int_{0}^{\infty}\frac{s^{a-1}}{(s+1)^{a+b}}ds$$
If one now follows my method and substitute b=1-a, one gets the rather simple integral
$$B(a,1-a)=\int_{0}^{\infty}\frac{s^{a-1}}{s+1}ds$$

In this step, contour integration over the complex plane seems to work. After a couple tries, I found a complex-valued function that requires the evaluation of this integral in its complex line integral. I first chose the contour C as a full circle around the origin that excludes the negative real axis in the fashion of a keyhole contour. Then, I picked the complex valued function

$$\int_{C}\frac{z^{a-1}}{1-z}dz$$

where C is the contour just specified. I then assume that $0\leq \Re z\leq 1$. (This assumption can be made without loss of generality, because through analytic continuation, the properties of the Gamma function that hold in that strip holds for all values of z.)

I first evaluate the whole contour integral using the residue theorem. There exists only one pole of this complex valued function, and that is at z=1. The residue there can easily be obtained and is -1, and hence, from the residue theorem, follows the value of the contour integral:

$$\int_{C}\frac{z^{a-1}}{1-z}dz = -2\pi i$$

Now, we name the radius of the large circle R and the small circle r (the two parts of the contour in a keyhole contour). By the definition of the complex line integral, the contributions of the two circles are respectively: (after substituting in the parametrization of the circles)

$$\int_{-\pi}^{\pi}\frac{iR^xe^{ix\theta}}{1-Re^{i\theta}}d\theta$$
and
$$\int_{\pi}^{-\pi}\frac{ir^xe^{ix\theta}}{1-re^{i\theta}}d\theta$$

These integrals both vanish as R goes to infinity and r goes to zero. Hence, we are only left with the contribution of the two parts of the keyhole around the negative real axis. These parts are both parametrized by z=-t, and from the definition of the complex logarithm (the two parts of the remaining contour are in two different branchs) we have that

$$-2\pi i = -\int_{\infty}^{0}\frac{e^{i\pi(a-1)}t^{a-1}}{1+t}dt-\int_{0}^{\infty}\frac{e^{-i\pi (a-1)}t^{a-1}}{1+t}dt$$

Using some exponential properties, this simplifies to

$$-2\pi i = (e^{-i\pi a}-e^{i\pi a})\int_{0}^{\infty}\frac{t^{a-1}}{1+t}dt$$

and the desired result follows from Euler's identity:

$$\Gamma(a)\Gamma(1-a)=\int_{0}^{\infty}\frac{t^{a-1}}{1+t}dt=\frac{\pi}{\sin(\pi a)}$$

Can anybody check this for errors?

Last edited: Jul 7, 2012