TITLE: what is solution of these improper integrals?
QUESTION [7 upvotes]: $$\int_{0}^{\infty} x^{n}\sinh x dx$$
$$\int_{0}^{\infty} x^{n}\cosh x dx$$
$$\int_{0}^{\infty} x^{n}\tanh x dx$$
what is solution of these improper integrals?

REPLY [3 votes]: Recalling the Mellin transform of a function $ f(x) $,   
$$ F(s) = \int_{0}^{\infty} x^{s-1}f(x)\, dx \,.$$ 
Then one can write 
$$ \int_{0}^{\infty} x^{n}\tanh(x)\, dx = F(n+1) \,, $$
where $ F(n) $ is the Mellin transform of $\tanh(x)$.
The Mellin transform of $\tanh(x)$ (computed by Maple) is given by 
$$ F(s) = {2}^{1-s} \left( {2}^{1-s}-1 \right) \Gamma  \left( s \right) \zeta \left( s \right)\,, \quad -1<\Re(s)<0 .$$ 
Our integral can be evaluated as
$$ \int_{0}^{\infty} x^{n} \tanh(x)\, dx  = {2}^{-n} \left( {2}^{-n}-1 \right) \Gamma  \left( n+1 \right) \zeta \left( n+1 \right)\,. $$
The above integral exists for $ -2 < n < -1  $. Here is a comparison between the numerical evaluation of the integral for $n=-\frac{3}{2}$ and the one computed by the formula for the same $n$
Numerical evaluation of the integral $ \approx 3.811125882 $
$$ -4\,\sqrt {2} \left( 2\,\sqrt {2}-1 \right) \sqrt {\pi }\zeta \left( -\frac{1}{2} \right) = 3.811125880 $$.
See the gamma and zeta functions.