# 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.

• Nice question! Haven't figured it out yet... Aug 31 '21 at 23:56
• I'm also interested by a formal explanation over this... Sep 6 '21 at 0:39
• If it can be proved that derivation of the above integral w.r.t $\alpha$ is zero then I think it would be a nice solution! Sep 7 '21 at 11:28
• If the question can’t be solved here you could always try mathoverflow. Sep 12 '21 at 0:15

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

• Thanks. BTW $\alpha=\pi n$ for any positive integer $n$ can be solved too converting it to a finite sum of integrals $\int_0^\infty \frac{\cosh(\pi k x)}{\cosh(\pi nx)} \sin (\pi n x^2)\, dx$, where each integral can be calculated by known formulas as a finite sum. Then if the resulting double sum can be verified independently, this will give a proof for $\alpha=\pi n$. Then one might try to use en.wikipedia.org/wiki/Carlson%27s_theorem or some of its analogs. However the problem here is the integral is not an analytic function of $\alpha$. Sep 12 '21 at 7:55
• @Nemo I initially tried to exploit the fact that if $$f(z) = \frac{\tanh( n \pi z)}{\tanh(\pi z)} \frac{e^{i 2n\pi (z^{2}+1)}}{2\sinh(4 n \pi z)},$$ then $f(x-i) - f(x+i) = \frac{\tanh(n \pi z)}{\tanh(\pi z)} e^{i 2 n \pi x^{2}}$. Sep 12 '21 at 15:16