Two Mathematics Stack Exchange Questions and Answers on the Gamma Function

Publicado em 6 de Maio de 2012 por

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.

Anúncios

Sobre Américo Tavares

eng. electrotécnico reformado / retired electrical engineer
Esta entrada foi publicada em Cálculo, Função Gama, Funções Especiais, Integrais, Integrais impróprios, Matemática, Math, Mathematics Stack Exchange com as etiquetas , , . ligação permanente.

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