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I have numerically computed the integral $\int\limits_0^{+\infty}\frac{x\pi}{x\pi+2\sinh(x\pi)} \, dx$ such that it's value is a rational number and it's equal $0.298549$. An inverse symbolic calculator doesn't give anything. I think that it may have a closed form since it's related to the exponential function. How can I evaluate that in a closed form?

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    $\begingroup$ Mathematica does not evaluate it. $\endgroup$ – Julián Aguirre Jul 3 '18 at 18:29
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    $\begingroup$ In Wolfram Alpha I got the same result OP stated, but as far as I know, WA truncates the result. It might actually be an irrational number. SageMath does not evaluate that integral, either. $\endgroup$ – Ivo Terek Jul 3 '18 at 18:32
  • $\begingroup$ @IvoTerek If you make the substitution $u=x\pi$ then WA does not truncate it: see here under Definite Integral. $\endgroup$ – TheSimpliFire Jul 3 '18 at 18:48
  • $\begingroup$ $$ [0; 3, 2, 1, 6, 5, 4, 2, 2, 1, 1, 1, 4, 1, 4, 2, 1, 4, 1, 2,\ldots]$$ does not appear to be a rational number and you can probably prove it by exploiting the integrals $\int_{\mathbb{R}}\left(\frac{\pi x}{\sinh(\pi x)}\right)^m\,dx.$ $\endgroup$ – Jack D'Aurizio Jul 3 '18 at 19:09
  • $\begingroup$ look Wolfram alpha in this case dosn't trancate it $\endgroup$ – zeraoulia rafik Jul 3 '18 at 21:14
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$$\frac{1}{\pi}\int_{0}^{+\infty}\frac{z\,dz}{z+2\sinh z}=\frac{1}{\pi}\int_{0} ^{1}\frac{\log u}{u\log u+u^2-1}\,du=\frac{1}{2\pi}\int_{\mathbb{R}}\frac{z\,dz}{z+2\sinh(z)}. $$ Let $\zeta_1,\zeta_2,\zeta_3,\ldots$ be the sequence of points in the first quadrant where $z+2\sinh z$ equals zero, ordered according to their distance from the origin. These points, together with $-\overline{\zeta_1},-\overline{\zeta_2}$ etcetera, are the simple poles of $\frac{z}{z+2\sinh(z)}$ in the upper-half plane. We have

$$ \operatorname*{Res}_{z=\zeta_i} \frac{z}{z+2\sinh(z)}=\lim_{z\to \zeta_i}\frac{z(z-\zeta_i)}{z+2\sinh(z)}\stackrel{\text{d.H.}}{=}\frac{\zeta_i}{1+2\cosh(\zeta_i)}=\frac{\zeta_i}{1\pm \sqrt{4+\zeta_i^2}}\tag{A}$$ hence the sum of the residues at $\zeta_i$ and $-\overline{\zeta_i}$ is given by $$ \frac{ai(-4\sinh a\sin b)+bi(2+4\cosh a\cos b)}{(1+2\cosh a\cos b)^2+(2\sinh a \sin b)^2},\quad a=\text{Re}(\zeta_i), b=\text{Im}(\zeta_i) $$ where $ \zeta_1 \approx 1.48748 + 4.35037 i, \zeta_2 \approx 2.39367 + 10.7734 i, \zeta_3\approx 2.85034 + 17.1126 i$.
An approximate location of the roots of $z+2\sinh z$ leads to an approximate integral, which can be computed through standard methods applied to $\frac{1}{\pi}\int_{0}^{1}\frac{\log u}{u\log u+u^2-1}\,du$.

It is interesting to point out that if the determination of $\sqrt{\cdot}$ in $(A)$ were always the same, the integral would have a closed form in terms of the coefficients of the Maclaurin series of an algebraic/trigonometric function. Unluckily, $\zeta_1$ is associated to $-$, $\zeta_2$ is associated to $-$, $\zeta_3$ is associated to $+$, $\zeta_4$ is associated to $-$...

On the other hand the fact that Mathematica does not recognize a closed form for such integral does not rule out the chance to apply to the given integral an approach similar to the one used for evaluating $\int_{\mathbb{R}}\frac{dx}{(x-e^x)^2+\pi^2}$.

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