Two Mathematics Stack Exchange Questions and Answers on the Gamma Function

I post here together two answers of mine on MSE both on the gamma function.

Question by pomme. Definition of the gamma function

“I know that the Gamma function with argument $(-\dfrac{1}{ 2})$ — in other words $\Gamma(-\dfrac{1}{2})$ is equal to $-2\pi^{1/2}$ . However, the definition of

$\Gamma(k)=\displaystyle\int_0^\infty t^{k-1}e^{-t}dt$

but how can $\Gamma(-\frac{1}{2})$ be obtained from the definition? WA says it does not converge…”

My answer

Your doubt makes sense if for $k=-1/2$ you try using the definition of the gamma function by the integral you have written, because it diverges at $k=-1/2$ as you stated. I will try to clarify if as follows. This integral representation of the gamma function (I use $x$ instead of $k$ )

$\Gamma (x)=\displaystyle\int_{0}^{\infty }t^{x-1}e^{-t}dt\qquad(0)$

holds in the reals if and only if $x>0$ . Using integration by parts we can show that

$\Gamma (x+1)=x\Gamma (x).\qquad(1)$

For $x<0$ we can define $\Gamma (x)$ for all negative values of $x<0$ except $-1,-2,-3,\cdots$ not by the integral $(0)$ , rather by means of the functional equation $(1)$ in the form

$\Gamma (x)=\dfrac{\Gamma(x+1)}{x}.\qquad(2)$

Then $x+1>0$ [and] $\Gamma (x+1)$ is convergent. In your example $x=-1/2$ , so $x+1=1/2$ and $\Gamma (1+x)=\Gamma (1/2)$ . Hence we obtain

$\Gamma (-1/2)=\dfrac{\Gamma(1/2)}{-1/2}=-2\Gamma (1/2)=-2\sqrt{\pi },\qquad(3)$

where we use the known value of the integral $\Gamma(1/2)=\sqrt{\pi }$ , which can be evaluated, e.g. from the equality

$\displaystyle\int_{0}^{\infty }\displaystyle\int_{0}^{\infty }e^{-x^{2}-y^{2}}dxdy=\dfrac{\pi }{4}=\left( \displaystyle\int_{0}^{\infty }e^{-x^{2}}dx\right) ^{2}.\qquad(4)$

Similarly, for $x=-3/2$ by $(2)$ we find $\Gamma (-3/2)=-\frac{2}{3}\Gamma (-1/2)$ , using $(3)$ twice. This process is called analytic continuation, but its true understanding requires knowledge of complex analysis.

$\text{Plot of }y=\Gamma(x)\quad -5<x<5$

Question by Amitabh Udayiman. Convergence of this integral

“My statistics text book prescribed by my school states that the integral

$\Gamma(n)=\displaystyle\int_{0}^{\infty}e^{-x}x^{n-1}dx$

is convergent for $n>0$ .It does not prove the statement. So can anyone please help me prove it? Thanks again!”

My answer. I assume that $n$ is a real number. Split the gamma improper integral

$\Gamma(n)=\displaystyle\int_{0}^{\infty}e^{-x}x^{n-1}dx\qquad(0)$

into $I_1+I_2$ , where

$I_1=\displaystyle\int_{0}^{1}e^{-x}x^{n-1}dx\qquad(1)$

and

$I_2=\displaystyle\int_{1}^{\infty}e^{-x}x^{n-1}dx\qquad(2)$

1. To prove that the integral $I_2$ is always convergent use the fact that for any real number $\alpha$ the integral

$\int_{1}^{\infty }e^{-x}x^{\alpha }dx\qquad(3)$

is convergent, by the limit comparison test

$\displaystyle\lim_{x\rightarrow \infty }\dfrac{e^{-x}x^{\alpha }}{x^{-2}}=0\qquad(4)$

with the convergent integral

$\displaystyle\int_{1}^{\infty }\dfrac{dx}{x^{2}}.\qquad(5)$

2. As for $I_1$ consider two cases. (a) If $n\geq 1$ observe that $\lim_{x\rightarrow 0}e^{-x}x^{n-1}=0$ , so $I_1$ is a proper integral. (b) If $0<n<1$ , the integrand $e^{-x}x^{n-1}$ behaves like $x^{n-1}$ near $n=0$ , because $e^{-x}\rightarrow 1$ as $x\rightarrow 0$ . Since

$\displaystyle\int_{0}^{1}\dfrac{dx}{x^{1-n}}\qquad(6)$

is convergent if and only if $1-n<1$ , i.e. $n>0$ , so is $I_1$ . It follows that $\Gamma(n)=I_1+I_2$ is convergent for $n>0.$

Sobre Américo Tavares

eng. electrotécnico reformado / retired electrical engineer

Ver todos os artigos de Américo Tavares →

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