Friend claims $\int_0^\infty\sum_{n=0}^{\infty}\frac{x^n}{2^{(n+1)^sx^{n+1}}+1}dx=\zeta(s+1)$?

Question

My friend is making another claim on another integral!

Can anybody verify it? Or his is mocking on me?

Valid for all $s\ge1$

$$\int_0^\infty\sum_{n=0}^{\infty}\frac{x^n}{2^{(n+1)^sx^{n+1}}+1}dx=\zeta(s+1)$$

Where $\zeta(s)$ is the Zeta function defined for all $R(s)>1$

$$\zeta(s)=\sum_{n=0}^{\infty}\frac{1}{(n+1)^s}$$

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I think this integral is fault because, most integral with $\zeta(s)$ involves an $e^x$ at the denominator or a $\ln(x)$. This integral has none of these. I also check maths world and other sites but couldn't an integral similar to it.

Answer 1

Hint. One may just interchange sum and integration : $$ \int_0^\infty\sum_{n=0}^{\infty}\frac{x^n}{2^{(n+1)^sx^{n+1}}+1}dx=\sum_{n=0}^{\infty}\int_0^\infty\frac{x^n}{2^{(n+1)^sx^{n+1}}+1}dx $$ then, by the change of variable $$u=(n+1)^sx^{n+1}, \quad du=(n+1)^{s+1}x^ndx,$$ one gets $$ \int_0^\infty\frac{x^n}{2^{(n+1)^sx^{n+1}}+1}dx=\frac1{(n+1)^{s+1}}\int_0^\infty\frac1{2^u+1}du=\frac1{(n+1)^{s+1}} \times 1 =\frac1{(n+1)^{s+1}}, $$ giving

$$ \int_0^\infty\sum_{n=0}^{\infty}\frac{x^n}{2^{(n+1)^sx^{n+1}}+1}dx=\zeta(s+1),\quad s\geq1, $$

as announced.