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This question already has an answer here:

$$ I=\displaystyle \int_{0}^{\infty} \dfrac{\sin (\pi x^{2})}{\sinh^{2} (\pi x)} ~\mathrm{d}x $$ 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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marked as duplicate by N3buchadnezzar, tired, Nosrati, JonMark Perry, carmichael561 Apr 7 '17 at 1:20

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ The language is Norwegian. $\endgroup$ – Connor Harris Apr 6 '17 at 18:43
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    $\begingroup$ Kokeboken = cookbook? $\endgroup$ – Omnomnomnom Apr 6 '17 at 18:45
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    $\begingroup$ Some insight: math.stackexchange.com/questions/2096675/… $\endgroup$ – Ron Gordon Apr 6 '17 at 18:47
  • $\begingroup$ 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. $\endgroup$ – Jeff Strom Apr 6 '17 at 18:59
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    $\begingroup$ Oh, hi. That's mine ^^ $\endgroup$ – N3buchadnezzar Apr 6 '17 at 20:07
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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.

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