TITLE: An integral related with the Riemann $\zeta$ function
QUESTION [6 upvotes]: I have to prove that:
$$
\forall s>1,\qquad\int_0^\infty \sum_{k=1}^{\infty}\frac{1}{(k^s+1)^x+k^s}dx=\zeta(s).
$$
I how do I find the closed form for this sum?
$$
\sum_{k=1}^{\infty}\frac{1}{(k^s+1)^x+k^s}
$$

REPLY [3 votes]: Just to complete Ron Gordon's excellent answer, notice that:
$$ I(a,b)=\int_{0}^{+\infty}\frac{dx}{e^{x\log a}+b} = \frac{1}{\log a}\int_{0}^{+\infty}\frac{dz}{e^z+b}=\frac{1}{\log a}\int_{1}^{+\infty}\frac{dt}{t(t+b)}$$
and:
$$ \int_{1}^{+\infty}\frac{dt}{t(t+b)}=\int_{0}^{1}\frac{du}{1+bu}=\frac{1}{b}\int_{0}^{b}\frac{dv}{1+v}=\frac{\log(1+b)}{b}$$
through the substitutions $x=\frac{z}{\log a}$, $z=\log t$, $t=\frac{1}{u}$ and $u=\frac{v}{b}$. That leads to:
$$\int_{0}^{+\infty}\sum_{k\geq 1}\frac{dx}{(k^s+1)^x+k^s}=\sum_{k\geq 1}\frac{\log(1+k^s)}{k^s \log(1+k^s)} = \zeta(s)$$
as wanted.