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$$\int_{0}^{\infty} x^{n}\sinh x dx$$

$$\int_{0}^{\infty} x^{n}\cosh x dx$$

$$\int_{0}^{\infty} x^{n}\tanh x dx$$

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 $2012-09-02
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    Ehh. What have you tried?2012-09-02
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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.$2012-09-02
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    This question does not show an ounce of thinking. -1.2012-09-02
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    @did cooooool ok ill turn it back2012-09-02
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    What do you call *to turn it back*?2012-09-02
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    @albert2: Your third integral can have a closed form solution and exists for $-2. See the answer.2012-09-02

3 Answers 3

9

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? Mhenni seems to make it not diverge there.2014-08-31
3

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}

3

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 $-1<\Re(s)<0$.2012-09-02
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    @OP Is this supposed to answer the question?2012-09-02
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    @Sasha: Are you reminding me about the strip of convergence? Thank you.2012-09-02
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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.2012-09-02
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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.2012-09-02
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    @Downvoter:This is an evaluation of the integral. Try to work it out.2012-09-02
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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](http://en.wikipedia.org/wiki/Reciprocal_Gamma_function#Contour_integral_representation) of the $\Gamma$-function. The question of analytic continuation is not the question OP had asked.2012-09-02
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    @Saash:I see what you mean, I'll include it. Thank you.2012-09-02
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    @OP:This is an answer to your third integral.2012-09-02