More direct proof that $ \int_0^\infty\frac{\tanh(a x)}{\tanh(\pi x)}\sin(2ax^2)\,dx=\frac{1}{4}$ is independent of parameter $a$

Question

I'm looking for alternative ways to calculate the integral $$ \int_0^\infty\frac{\tanh(\alpha x)}{\tanh(\pi x)}\sin(2\alpha x^2)\,dx=\frac{1}{4},\qquad \alpha>0.\tag {*} $$

It was derived in a lengthy calculation using roundabout method from Fourier transform of a function of two variables. However, it seems like the integral (*) might have a simple proof? Note that when $\alpha=\pi$, the the integral reduces to Fresnel integral $$ \int\limits_0^\infty\sin(2\pi x^2)\,dx=\frac{1}{4}. $$ Thus if one could prove that it is independent of the parameter $\alpha$, then its value could be found.

Answer 1

Since no one has posted an answer, I'm going to use contour integration to show that the equation holds for the next "simplest" case, namely $\alpha = 2 \pi$.

$$\begin{align} \int_0^\infty \frac{\tanh(2 \pi x)}{\tanh(\pi x)} \sin (4 \pi x^2) \, \mathrm dx &= \int_{0}^{\infty} \frac{2 \cosh^2(\pi x)}{\cosh (2 \pi x)} \sin (4 \pi x^2) \, \mathrm dx\\ &= \int_0^\infty \left(1+ \frac{1}{\cosh (2 \pi x)} \right) \sin(4 \pi x^{2}) \, \mathrm dx \\ &= \frac{1}{2} \int_0^\infty \left(1+ \frac{1}{\cosh (\pi u)} \right) \sin (\pi u^2) \, \mathrm d u \\ &= \frac{1}{2}\left( \int_0^\infty \sin (\pi u^2) \, \mathrm du + \int_0^\infty \frac{\sin(\pi u^2)}{\cosh(\pi u)} \, \mathrm du \right) \\ &= \frac{1}{2} \left(\frac{\sqrt{2}}{4} + \int_0^\infty \frac{\sin(\pi u^2)}{\cosh(\pi u)} \, \mathrm du\right) \end{align}$$

To evaluate the integral $$\int_0^\infty \frac{\sin(\pi u^2)}{\cosh(\pi u)} \, \mathrm du $$ we can integrate the complex function $$f(z) = -\frac{e^{i \pi (z^{2}+1/4)}}{\sinh(2 \pi z)}$$ counterclockwise around an indented rectangular contour with vertices at $\pm R \pm \frac{i}{2} $.

As $R \to \infty$, the integral vanishes on the left and right sides of the rectangle since the hyperbolic sine function grows exponentially as $\Re(z) \to \pm \infty$.

And since $$f \left(x- \frac{i}{2} \right)- f \left(x+ \frac{i}{2} \right) = \frac{ e^{i \pi x^{2}}}{\cosh (\pi x)}, $$ we get

$$ \begin{align} \int_{-\infty}^\infty \frac{e^{i \pi x^2}}{\cosh(\pi x)} \, \mathrm dx &= \pi i \operatorname{Res}\left[f(z), -i/2\right] + 2 \pi i \operatorname{Res}[f(z), 0] + \pi i \operatorname{Res}[f(z), i/2] \\ &= \pi i \left(\frac{1}{2 \pi} \right) + 2 \pi i \left(- \frac{\sqrt{2}}{4 \pi} (1+i) \right) + \pi i \left(\frac{1}{2 \pi} \right) \\ &= \frac{\sqrt{2}}{2} + i \left( 1- \frac{\sqrt{2}}{2}\right). \end{align}$$

Therefore, $$\int_0^\infty \frac{\sin(\pi x^2)}{\cosh(\pi x)} \, \mathrm dx = \frac{1}{2}- \frac{\sqrt{2}}{4}, $$ and $$\int_0^\infty \frac{\tanh(2 \pi x)}{\tanh(\pi x)} \sin (4 \pi x^2) \, \mathrm dx = \frac{1}{2} \left(\frac{\sqrt{2}}{4} + \frac{1}{2} - \frac{\sqrt{2}}{4} \right) = \frac{1}{4}. $$

Answer 2

I was able to evaluate a closely related integral $$ I=\int\limits_0^\infty\frac{\coth(\alpha x)}{\sinh(\pi x)}\sin(\alpha x^2)\,dx=\frac{1}{4},\qquad \alpha>0, $$ using the integral $$ \int\limits_0^\infty\frac{\cos(at)}{\cosh t+\cosh b}\,dt=\frac{\pi\sin (ab)}{\sinh b\sinh(\pi a)}, $$ as follows. We write $$ \frac{\coth(\alpha x)}{\sinh(\pi x)}\sin(\alpha x^2)=\frac{1}{\pi}\int\limits_0^\infty\frac{\cos(xt)\cosh (\alpha x)}{\cosh t+\cosh(\alpha x)}\,dt, $$ and obtain $$ I=\frac{1}{\pi}\int\limits_0^\infty\int\limits_0^\infty\frac{\cos(xt)\cosh (\alpha x)}{\cosh t+\cosh(\alpha x)}\,dtdx=\frac{1}{4\pi}\int\limits_{-\infty}^\infty\int\limits_{-\infty}^\infty\frac{\cos(xt)\cosh (\alpha x)}{\cosh t+\cosh(\alpha x)}\,dtdx. $$ After change of variables $$ t=u+v,\quad \alpha x=u-v, $$ the integral becomes $$ I=\frac{1}{4\pi}\int\limits_{-\infty}^\infty\int\limits_{-\infty}^\infty\frac{\cos\frac{u^2-v^2}{\alpha}}{2\cosh (u)\cosh(v)}\left(\cosh (u)\cosh (v)-\sinh(u)\sinh(v)\right)\,\frac{2dudv}{\alpha}. $$ The odd terms containing $\sinh(u)\sinh(v)$ cancel after integration. After this, $\cosh (u)\cosh (v)$ in the numerator and denominator cancel out. The resulting integral is reduced to Fresnel integrals and easily evaluated.