Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How would you proof that $$I:=\int_{0}^{\infty}\frac{z^{x-1}}{e^{z}+1}dz=\left(1-2^{1-x}\right)\Gamma(x)\zeta(x)$$ I can rewrite the integral as $$I=\int_{0}^{\infty}z^{x-1}e^{-z}\sum_{n=0}^{\infty}(-1)^{n}e^{-nz}dz$$ but then I get stuck. Can you help me?

share|improve this question
2  
Integrate term by term. You should recognize $\left(1-2^{1-x}\right)\zeta(x)$ as $\sum_n (-1)^{n+1}n^{-x}$. –  stopple Mar 1 '13 at 0:03
2  
(Interchange the order of summation and integration.) –  anon Mar 1 '13 at 0:26
1  
A related problem. –  Mhenni Benghorbal Mar 1 '13 at 1:01

1 Answer 1

up vote 1 down vote accepted

It turns out that it's not so hard. I already had $$I=\int_{0}^{\infty}z^{x-1}e^{-z}\sum_{n=0}^{\infty}(-1)^{n}e^{-nz}dz,$$ so, interchanging the summation and integration like @anon says: (actually I don't know what criteria am I using but it results that: ) $$I=\sum_{n=0}^{\infty}(-1)^{n}\int_{0}^{\infty}z^{x-1}e^{-(n+1)z}dz.$$ Now, like @Mhenni Benghorbal suggested, let $y:=(n+1)z$, then $$I=\sum_{n=0}^{\infty}(-1)^{n}\int_{0}^{\infty}\left(\frac{y}{n+1}\right)^{x-1}e^{-y}\frac{1}{n+1}dy=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{\left(n+1\right)^{x}}\int_{0}^{\infty}y^{x-1}e^{-y}dy,$$ or $$I=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{x}}\Gamma(x).$$ But, separating the even and odd powers we have $$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{x}}=\sum_{n=1}^{\infty}\frac{(-1)^{2n-2}}{(2n-1)^{x}}+\sum_{n=1}^{\infty}\frac{(-1)^{2n-1}}{(2n)^{x}}=\sum_{n=1}^{\infty}\frac{1}{(2n-1)^{x}}-\sum_{n=1}^{\infty}\frac{1}{(2n)^{x}},$$ and using the fact that: (separating the even and odd powers) $$\sum_{n=1}^{\infty}\frac{1}{n^{x}}=\sum_{n=1}^{\infty}\frac{1}{(2n-1)^{x}}+\sum_{n=1}^{\infty}\frac{1}{(2n)^{x}},$$ then $$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{x}} = \left[\sum_{n=1}^{\infty}\frac{1}{n^{x}}-\sum_{n=1}^{\infty}\frac{1}{(2n)^{x}}\right]-\sum_{n=1}^{\infty}\frac{1}{(2n)^{x}},$$ or $$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{x}}=\sum_{n=1}^{\infty}\frac{1}{n^{x}}-2^{1-x}\sum_{n=1}^{\infty}\frac{1}{n^{x}}=\left(1-2^{1-x}\right)\zeta(x),$$ and finally $$I=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{x}}\Gamma(x)=\left(1-2^{1-x}\right)\zeta(x)\Gamma(x).$$

Thanks for your help! And if you know about the criteria of interchanging the sum and integration, please tell me. Also if you find an error.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.