Calculating an almost Gamma integral 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?
 A: 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.
