# 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?

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This is a funny question (+1). For instance, by AM-GM you have that $\int_0^\infty x^n=\left[\frac{x^{n+1}}{n+1}\right]_{0}^{\infty}\longrightarrow \infty\le \int_0^\infty x^n \frac{e^{-x}+e^{x}}{2} \ dx=\int_0^\infty x^n \cosh \ dx$ – I'm an artist Sep 2 '12 at 9:12
Ehh. What have you tried? – Jyrki Lahtonen Sep 2 '12 at 9:14
Maybe it also helps you that $\lim_{x\to\infty} \cosh x - \sinh x = \lim_{x\to\infty} e^{-x} = 0.$ – I'm an artist Sep 2 '12 at 9:40
This question does not show an ounce of thinking. -1. – Did Sep 2 '12 at 11:22
What do you call to turn it back? – Did Sep 2 '12 at 11:37

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.

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What if $-2<n<-1$? Mhenni seems to make it not diverge there. – Akiva Weinberger Aug 31 '14 at 18:04

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}

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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.

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The strip of converge of the Mellin transform of $\tanh(x)$ is $-1<\Re(s)<0$. – Sasha Sep 2 '12 at 12:45
@OP Is this supposed to answer the question? – Did Sep 2 '12 at 12:50
@Sasha: Are you reminding me about the strip of convergence? Thank you. – Mhenni Benghorbal Sep 2 '12 at 13:00
I hinting that $\int_0^\infty x^n \tanh(x) \mathrm{d}x$ diverges outside the strip. It may be well worth adding this to your post. – Sasha Sep 2 '12 at 13:06
@Sasha:The Mellin transform of ${\rm e}^{-x}$ is $\Gamma(s)$ with the strip of convergence $R(s) > 0$. Does that mean that the $\Gamma(s)$ does not exist out side the strip of convergence? Think about it for a while. – Mhenni Benghorbal Sep 2 '12 at 13:12