Ramanujan: $\sqrt{a}\int_0^\infty\frac{e^{-x^2}}{\cosh ax} dx=\sqrt{b}\int_0^\infty\frac{e^{-x^2}}{\cosh bx} dx$, where $ab=\pi$

Question

I found this integral in Ramanujan notebook. How to prove this?

If $\alpha \beta=\pi$, show that $$\sqrt{\alpha} \int_{0}^{\infty} \frac{e^{-x^2}}{\cosh(\alpha x)} dx=\sqrt{\beta} \int_{0}^{\infty} \frac{e^{-x^2}}{\cosh(\beta x)} dx$$

Answer 1

Here is my approach by using two Fourier transform: $$\int_{-\infty}^{\infty} \frac{e^{-2 \pi i x \xi}}{\cosh(\pi x)} dx= \frac{1}{\cosh ( \pi \xi)};\space \int_{-\infty}^{\infty} e^{-\pi x^2}e^{-2\pi i x \xi}dx = e^{-\pi \xi^2}$$ Notice that the functions inside the integrand are both even, one has: $$LHS = \frac{\sqrt{a}}{2}\int_{-\infty}^{\infty}\frac{e^{-x^2}}{\cosh(\alpha x)} dx \overset{y = \frac{ax}{\sqrt{\pi}}} = \frac{\sqrt{\pi}}{2\sqrt{a}} \int_{-\infty}^{\infty}\frac{e^{-\pi \frac{y^2}{a^2}}}{\cosh\left(\sqrt{\pi}y\right)} dy$$ $$= \frac{\sqrt \pi}{2\sqrt{a}} \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{e^{- \pi t^2}e^{-2\pi i t\cdot \frac{y}{a}}}{\cosh\left(\sqrt{\pi}y\right)} dtdy = \frac{\sqrt \pi}{2\sqrt{a}} \int_{-\infty}^{\infty}e^{- \pi t^2}\int_{-\infty}^{\infty}\frac{e^{-2\pi it \cdot \frac{y}{a}}}{\cosh\left(\sqrt{\pi}y\right)} dydt$$ $$\overset{y =\sqrt\pi z}=\frac{\pi}{2\sqrt{a}} \int_{-\infty}^{\infty}e^{-\pi t^2}\int_{-\infty}^{\infty} \frac{e^{-2\pi i t\cdot \frac{\sqrt \pi z}{a}}}{\cosh(\pi z)} dzdt =\frac{\pi}{2\sqrt{a}} \int_{-\infty}^{\infty}\frac{e^{-\pi t^2}}{\cosh\left (\pi \frac{\sqrt \pi t}{a}\right )}dt$$ $$\overset{u = \sqrt{\pi} t}= \frac{\sqrt{\pi}}{2\sqrt{a}}\int_{-\infty}^{\infty}\frac{e^{-u^2}}{\cosh(\beta u)}du= \sqrt \beta\int_{0}^{\infty}\frac{e^{-u^2}}{\cosh(\beta u)}du = RHS $$