The integral $$I(r)=\int_0^{\infty} \frac{e^{-rx^2}}{\cosh x}\,dx$$ has no closed-form expression, however there is an exact identity relating small and large values of $r$: $$I(r)=\frac{1}{\sqrt{\pi r}}I\left(\frac{1}{\pi^2 r}\right),$$ with $I(0)=\pi/2$, $\lim_{r\rightarrow\infty}\sqrt{r}I(r)=\tfrac{1}{2}\sqrt{\pi}$ as special cases.
This follows from the fact that the cosine transforms ${F}(k)=\sqrt{2/\pi}\int_0^\infty f(x)\cos kx\,dx$ of $f(x)=e^{-x^2/2}$ and $g(x)=1/\cosh(\sqrt{\pi/2}x)$ are the same functions: ${F}(k)=f(k)$ and ${G}(k)=g(k)$. Substitution of the transform and exchange of the order of integration then gives $$\int_0^\infty f(x)g(\alpha x)\,dx=\int_0^\infty f(x){G}(\alpha x)\,dx=\int_0^\infty f(x)\left(\sqrt{2/\pi}\int_0^\infty g(y)\cos (\alpha xy)\,dy\right)\,dx$$ $$=\int_0^\infty F(\alpha y)g(y)\,dy=\int_0^\infty f(\alpha x)g(x)\,dx,$$ from which the identity follows.
Ramanujan gave this result in a letter to Hardy, in the symmetric form $$\sqrt{\alpha} \int_{0}^{\infty} \frac{e^{-x^2}}{\cosh\alpha x} dx=\sqrt{\beta} \int_{0}^{\infty} \frac{e^{-x^2}}{\cosh\beta x} dx,\;\;\text{if}\;\;\alpha\beta=\pi.$$ Hardy wrote back that he knew of the result and had published it.