$$ I=\displaystyle \int_{0}^{\infty} \dfrac{\sin (\pi x^{2})}{\sinh^{2} (\pi x)} ~\mathrm{d}x $$ Found this on the cover of a book called "Integral Kokeboken" written in some language that i've never seen and btw the ans is $$ I=\dfrac{2-\sqrt{2}}{4}$$
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5The language is Norwegian. – Connor Harris Apr 06 '17 at 18:43
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8Kokeboken = cookbook? – Ben Grossmann Apr 06 '17 at 18:45
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3Some insight: http://math.stackexchange.com/questions/2096675/does-int-0-infty-sin-pix2-over-sinh-pix-tanhx-pi-mathrm/2096812#2096812 – Ron Gordon Apr 06 '17 at 18:47
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Graph the integrand and you'll see it's similar to a gaussian with a peak around 0.319. It touches the x axis at x=1, and then the function seems to oscillate between positive and negative values as it settles to f(x)=0 at infinity. My point is that this value for the integral is not really too surprising when you look at the function. – Jeff Strom Apr 06 '17 at 18:59
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Okay i'll take a look at it @RonGordon – Siddhartha Apr 06 '17 at 19:07
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1Oh, hi. That's mine ^^ – N3buchadnezzar Apr 06 '17 at 20:07
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@N3buchadnezzar good catch – tired Apr 06 '17 at 22:07
1 Answers
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This integral follows a very similar solution path to the one to which I linked in the comments. Consider the following contour integral:
$$\oint_C dz \frac{\cos{\left (\pi z^2\right )}}{\sinh^3{\left (\pi z\right )} \cosh{\left (\pi z\right )}} $$
about the rectangle with vertices $\pm R \pm i$ with small semicircular detours around the poles at $z=\pm i$. The contour integral is then equal to
$$PV \int_{-R}^R dx \frac{\cos{[\pi (x-i)^2]}}{\sinh^3{[\pi (x-i)]} \cosh{[\pi (x-i)]}} \\+ i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{\cos{\left [\pi \left (-i+\epsilon e^{i \phi} \right )^2 \right ]}}{\sinh^3{\left [\pi \left (-i+\epsilon e^{i \phi} \right ) \right ]} \cosh{\left [\pi \left (-i+\epsilon e^{i \phi} \right ) \right ]}} \\ + PV \int_R^{-R} dx \frac{\cos{[\pi (x+i)^2]}}{\sinh^3{[\pi (x+i)]} \cosh{[\pi (x+i)]}} \\+ i \epsilon \int_{2 \pi}^{\pi} d\phi \, e^{i \phi} \frac{\cos{\left [\pi \left (i+\epsilon e^{i \phi} \right )^2 \right ]}}{\sinh^3{\left [\pi \left (i+\epsilon e^{i \phi} \right ) \right ]} \cosh{\left [\pi \left (i+\epsilon e^{i \phi} \right ) \right ]}} \\ + i \int_{-1}^1 dy \frac{\cos{\left [\pi \left (R+i y \right )^2 \right ]}}{\sinh^3{\left [\pi \left (R+i y \right ) \right ]} \cosh{\left [\pi \left (R+i y \right ) \right ]}}\\+i \int_1^{-1} dy \frac{\cos{\left [\pi \left (-R+i y \right )^2 \right ]}}{\sinh^3{\left [\pi \left (-R+i y \right ) \right ]} \cosh{\left [\pi \left (-R+i y \right ) \right ]}}$$
Note that the first and third integrals are actually expressed as Cauchy principal values because the individual integrals themselves do not converge. That said, when combined, the resulting integral does converge and we may remove the $PV$ label.
As $R \to \infty$, the last two integrals go to zero.
The second integral approaches, in the limit as $\epsilon \to 0$:
$$-i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{-1-2 \pi^2 \epsilon^2 e^{i 2 \phi}}{\left (\pi \epsilon e^{i \phi} \right )^3 \left (1 + \frac16 \pi^2 \epsilon^2 e^{i 2 \phi} + \cdots \right )^3 \left (1+\frac12 \pi^2 \epsilon^2 e^{i 2 \phi} \right )} \\ = -i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{-1-2 \pi^2 \epsilon^2 e^{i 2 \phi}}{\left (\pi \epsilon e^{i \phi} \right )^3} \left (1-\pi^2 \epsilon^2 e^{i 2 \phi} \right ) \to i$$
The fourth integral approaches an identical limit as $\epsilon \to 0$.
The first and third integrals combine to produce, as $R \to \infty$,
$$-i 4 \int_{-\infty}^{\infty} dx \frac{\sin{\left ( \pi x^2 \right )}}{\sinh^2{(\pi x)}} $$
The contour integral is also equal to $i 2 \pi$ times the sum of the residues of the integrand at the poles $z=0$ and $z=\pm i/2$. The residue at $z=0$ may be computed by expanding the integrand in a Laurent series about $z=0$, which is
$$\frac1{(\pi z)^3} \left (1 - \frac12 \pi^2 z^4+\cdots \right ) \left (1 - \frac12 \pi^2 z^2+\cdots \right )^2$$
The residue is the coefficient of $z^{-1}$, or $-1/\pi$. In a similar vein, the sum of the residues at the pole $z=\pm i/2$ is $\sqrt{2}/\pi$. Thus,
$$-i 4 \int_{-\infty}^{\infty} dx \frac{\sin{\left ( \pi x^2 \right )}}{\sinh^2{(\pi x)}} + i 2 = i \left ( 2 \sqrt{2}-2 \right )$$
Rearranging things a bit, we find that the original integral is
$$\int_0^{\infty} dx \frac{\sin{\left ( \pi x^2 \right )}}{\sinh^2{(\pi x)}} = \frac{2-\sqrt{2}}{4} $$
as asserted.
Ron Gordon
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