# Does $\int_{0}^{\infty}{\sin{(\pi{x^2})}\over \sinh{(\pi{x}})\tanh(x\pi)}\mathrm{d}x$ have a simple closed from?

Does this integral $(1)$ has a simple closed form?

$$\int_{0}^{\infty}{\sin{(\pi{x^2})}\over \sinh{(\pi{x}})\tanh(x\pi)}\mathrm dx\tag1$$

An attempt of approach

$$\int_{0}^{\infty}{\sin{(\pi{x^2})}\cosh{(x\pi)}\over \sinh{(\pi{x}})^2}\mathrm dx\tag2$$

$$\int_{0}^{\infty}{\sin{(\pi{x^2})}\cosh{(x\pi)}\over 1-\cosh{(\pi{x}})^2}\mathrm dx\tag3$$

$$\int_{0}^{\infty}{\sin{(\pi{x^2})}\cosh{(x\pi)}\over (1-\cosh{(\pi{x}}))(1+\cosh{(x\pi)})}\mathrm dx\tag4$$

$${1\over2}\int_{0}^{\infty}{\sin{(\pi{x^2})}\over 1-\cosh{(\pi{x}})}\mathrm dx-{1\over 2}\int_{0}^{\infty}{\sin{(\pi{x^2})}\over 1+\cosh{(\pi{x}})}\mathrm dx\tag5$$

Not sure what to do next. Any help?

• The value seems to be $\frac{1}{4}$ exactly Commented Jan 13, 2017 at 21:15
• math.stackexchange.com/questions/1987764/… should help Commented Jan 13, 2017 at 21:15

This integral may be evaluated using the residue theorem, by considering the complex integral

$$\oint_C dz \frac{\cos{\left (\pi z^2\right )}}{\sinh^3{\left (\pi z\right )}}$$

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

$$PV \int_{-R}^R dx \frac{\cos{[\pi (x-i)^2]}}{\sinh^3{[\pi (x-i)]}} + i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{\cos{\left [\pi \left (-i+\epsilon e^{i \phi} \right )^2 \right ]}}{\sinh^3{\left [\pi \left (-i+\epsilon e^{i \phi} \right ) \right ]}} \\ + PV \int_R^{-R} dx \frac{\cos{[\pi (x+i)^2]}}{\sinh^3{[\pi (x+i)]}} + i \epsilon \int_{2 \pi}^{\pi} d\phi \, e^{i \phi} \frac{\cos{\left [\pi \left (i+\epsilon e^{i \phi} \right )^2 \right ]}}{\sinh^3{\left [\pi \left (i+\epsilon e^{i \phi} \right ) \right ]}} \\ + i \int_{-1}^1 dy \frac{\cos{\left [\pi \left (R+i y \right )^2 \right ]}}{\sinh^3{\left [\pi \left (R+i y \right ) \right ]}}+i \int_1^{-1} dy \frac{\cos{\left [\pi \left (-R+i y \right )^2 \right ]}}{\sinh^3{\left [\pi \left (-R+i y \right ) \right ]}}$$

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 $\epsilon \to 0$:

$$i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{-1-2 \pi^2 \epsilon^2 e^{i 2 \phi}}{\left (\pi \epsilon e^{i \phi} \right )^3 \left (1 + \frac16 \pi^2 \epsilon^2 e^{i 2 \phi} + \cdots \right )^3} \\ = i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{-1-2 \pi^2 \epsilon^2 e^{i 2 \phi}}{\left (\pi \epsilon e^{i \phi} \right )^3} \left (1-\frac12 \pi^2 \epsilon^2 e^{i 2 \phi} \right ) \to -i \frac32$$

The fourth integral approaches an identical limit as $\epsilon \to 0$.

The first and third integrals combine to produce, as $R \to \infty$,

$$i 4 \int_{-\infty}^{\infty} dx \frac{\sin{\left ( \pi x^2 \right )}}{\sinh^2{(\pi x)}} \cosh{(\pi x)}$$

The contour integral is also equal to $i 2 \pi$ times the residue of the integrand at $z=0$. The residue may be computed by expanding the integrand in a Laurent series about $z=0$, which is

$$\frac1{(\pi z)^3} \left (1 - \frac12 \pi^2 z^4+\cdots \right ) \left (1 - \frac12 \pi^2 z^2+\cdots \right )$$

The residue is the coefficient of $z^{-1}$, or $-1/(2 \pi)$. Thus,

$$i 4 \int_{-\infty}^{\infty} dx \frac{\sin{\left ( \pi x^2 \right )}}{\sinh^2{(\pi x)}} \cosh{(\pi x)} - i 3 = -i$$

Rearranging things a bit, we find that the original integral is

$$\int_0^{\infty} dx \frac{\sin{\left ( \pi x^2 \right )}}{\sinh^2{(\pi x)}} \cosh{(\pi x)} = \frac14$$

I should mention a subtlety in evaluating the limit as $\epsilon \to 0$ above. The astute reader may have noticed that the dominant term in the integral could be singular as $\epsilon \to 0$, and would be if the result was an odd power in $\epsilon$. Nevertheless, as the power of $\epsilon$ is even, the contribution cancels upon integration.