show that $\int_0^\infty x^n\,\text{sech}^2x\,\mathrm dx=(2^{1-n}-4^{1-n})\,\Gamma(n+1)\,\zeta(n)$

Question

show that $$\int_0^\infty x^n\,\text{sech}^2x\,\mathrm dx=(2^{1-n}-4^{1-n})\,\Gamma(n+1)\,\zeta(n)$$

Answer 1

I will prove what I commented, namely that $$\int_0^\infty {\frac{{{x^{s - 1}}{e^x}}}{{{{\left( {1 + {e^x}} \right)}^2}}}{\text{d}}x} = {\mkern 1mu} \Gamma (s)\eta (s - 1)$$

Write $$\frac{{{e^{ - x}}}}{{{{\left( {1 + {e^{ - x}}} \right)}^2}}}$$ and use $$\frac{t}{{{{\left( {1 + t} \right)}^2}}} = \sum\limits_{n \geqslant 1} {{{\left( { - 1} \right)}^{n-1}}n{t^n}} $$

so that $$\frac{{{e^{ - x}}}}{{{{\left( {1 + {e^{ - x}}} \right)}^2}}} = \sum\limits_{n \geqslant 1} {{{\left( { - 1} \right)}^{n-1}}n{e^{ - nx}}} $$

Then $$\begin{align} \int\limits_0^\infty {\frac{{{x^{s - 1}}{e^x}}}{{{{\left( {1 + {e^x}} \right)}^2}}}dx} &= \sum\limits_{n \geqslant 1} {{{\left( { - 1} \right)}^n}n\int\limits_0^\infty {{e^{ - nx}}{x^{s - 1}}dx} } \cr &= \sum\limits_{n \geqslant 1} {{{\left( { - 1} \right)}^{n - 1}}\frac{1}{{{n^{s - 1}}}}\int\limits_0^\infty {{e^{ - x}}{x^{s - 1}}dx} } \cr &= \sum\limits_{n \geqslant 1} {{{\left( { - 1} \right)}^{n - 1}}\frac{1}{{{n^{s - 1}}}}\Gamma \left( s \right)} \cr &= \Gamma (s)\eta (s - 1) \end{align} $$

Try to see how to use this to get your integral (which I think is off by a factor or two or four).

Answer 2

Just use @ Peter Tamaroff's hint, $$\Gamma(s+1)\eta(s)=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^s}\int_0^\infty e^{-t}t^{s}\,dt=\int_0^\infty\sum_{n=1}^\infty (-1)^{n-1}ne^{-nt}t^{s}dt$$$$=\int_0^\infty\frac{e^{-t}t^s}{(e^{-t}+1)^2}\,dt=\int_0^\infty\frac{e^{t}t^s}{(e^{t}+1)^2}\,dt$$Substitute $t=2x$ and complete it.