Alternative and more direct proof that an integral is independent of a parameter I'm looking for alternative ways to calculate the integral
$$
\int\limits_0^\infty\frac{\tanh(\alpha x)}{\tanh(\pi x)}\sin(2\alpha x^2)\,dx=\frac{1}{4},\qquad \alpha>0.\tag {*}
$$
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.

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