what is solution of these improper integrals? $$\int_{0}^{\infty} x^{n}\sinh x dx$$
$$\int_{0}^{\infty} x^{n}\cosh x dx$$
$$\int_{0}^{\infty} x^{n}\tanh x dx$$
what is solution of these improper integrals?
 A: Since $\sinh x ,\cosh x\to \infty, \tanh x \to 1$ as $x \to \infty$, there is an $r>0$ such that 
$$
\sinh x \ge \frac{1}{2},\ \cosh x \ge \frac{1}{2},\ \tanh x \ge \frac{1}{2}
\forall x \ge r.
$$
It follows that
\begin{eqnarray}
\int_0^\infty x^n\sinh x dx&\ge&\int_r^\infty \frac{x^n}{2} dx=\infty,\cr
\int_0^\infty x^n\cosh x dx&\ge&\int_r^\infty \frac{x^n}{2} dx=\infty,\cr
\int_0^\infty x^n\tanh x dx&\ge&\int_r^\infty \frac{x^n}{2} dx=\infty.
\end{eqnarray}
A: 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.
A: All three of these hyperbolic functions are nonnegative when $x \geq 0.$ Thus,
$$\int_{0}^{\infty}x^n \sinh x dx > \int_{1}^{\infty}x^n \sinh x dx > \int_{1}^{\infty}\sinh x = \cosh x \bigg|_{1}^{\infty} = \frac{e^x + e^{-x}}{2} \bigg|_{1}^{\infty}$$
so that your first integral diverges to infinity.  For your second integral, interchange sinh and cosh above (and change the final plus sign to a minus sign accordingly). Again, the integral can be seen to diverge to infinity. Finally, recall (or check) that
$$ \int \tanh x dx = \log|\cosh x| + C$$
Bearing this in mind, it is clear that the third integral will also diverge to infinity.
In fact, if you simply write out what sinh, cosh, and tanh are in terms of the exponential function (i.e. their initial definitions), it should be clear that these indefinite integrals will all diverge to infinity.
