$$\int_{0}^{\infty} x^{n}\sinh x dx$$
$$\int_{0}^{\infty} x^{n}\cosh x dx$$
$$\int_{0}^{\infty} x^{n}\tanh x dx$$
what is solution of these improper integrals?
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4This 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 $ – user 1591719 Sep 02 '12 at 09:12
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5Ehh. What have you tried? – Jyrki Lahtonen Sep 02 '12 at 09:14
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Maybe it also helps you that $\lim_{x\to\infty} \cosh x - \sinh x = \lim_{x\to\infty} e^{-x} = 0.$ – user 1591719 Sep 02 '12 at 09:40
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6This question does not show an ounce of thinking. -1. – Did Sep 02 '12 at 11:22
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@did cooooool ok ill turn it back – albert2 Sep 02 '12 at 11:25
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2What do you call to turn it back? – Did Sep 02 '12 at 11:37
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@albert2: Your third integral can have a closed form solution and exists for $-2<n<-1$. See the answer. – Mhenni Benghorbal Sep 02 '12 at 13:47
3 Answers
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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.
Benjamin Dickman
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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
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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}
HorizonsMaths
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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 $ $$ -4\,\sqrt {2} \left( 2\,\sqrt {2}-1 \right) \sqrt {\pi }\zeta \left( -\frac{1}{2} \right) = 3.811125880 $$.
Mhenni Benghorbal
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The strip of converge of the Mellin transform of $\tanh(x)$ is $-1<\Re(s)<0$. – Sasha Sep 02 '12 at 12:45
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@Sasha: Are you reminding me about the strip of convergence? Thank you. – Mhenni Benghorbal Sep 02 '12 at 13:00
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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 02 '12 at 13:06
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@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 02 '12 at 13:12
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@Downvoter:This is an evaluation of the integral. Try to work it out. – Mhenni Benghorbal Sep 02 '12 at 13:16
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The Euler integral is divergent outside the strip of convergence. The $\Gamma(s)$ be can analytically continued by means of the functional equation, or using Hankel contour for the reciprocal of the $\Gamma$-function.
The question of analytic continuation is not the question OP had asked.
– Sasha Sep 02 '12 at 13:16 -
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