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 $r$: $$I(r)=\frac{1}{\sqrt{\pi r}}I\left(\frac{1}{\pi^2 r}\right).$$

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.

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.

Since the OP expressed in interest in Ramanujan's formula, this follows from the fact that the cosine transforms $\hat{F}(k)=\sqrt{2/\pi}\int_0^\infty f(x)\cos kx\,dx$ of $f(x)=e^{-x^2/2}$ and $g(x)=1/\cosh(\sqrt{\pi/2}x)$ are the same functions: $\hat{F}(k)=f(k)$ and $\hat{G}(k)=g(k)$. Substitution of the transform and exchange of the order of integration then gives $$\int_0^\infty f(x)g(\alpha x)\,dx=\int_0^\infty f(x)\hat{G}(\alpha x)\,dx=\int_0^\infty f(x)\left(\sqrt{2/\pi}\int_0^\infty g(y)\cos \alpha xy\,dy\right)\,dx$$ $$=\int_0^\infty f(\alpha y)g(y)\,dy,$$ from which the identity follows, $$\sqrt{\alpha} \int_{0}^{\infty} \frac{e^{-x^2}}{\cosh\alpha x} dx=\sqrt{\beta} \int_{0}^{\infty} \frac{e^{-x^2}}{\cosh\beta x} dx,\;\;\text{if}\;\;\alpha\beta=\pi.$$

Ramanujan gave this result in a letter to Hardy, without proof; Hardy wrote back that he knew of the result and had published it, so perhaps "Hardy's formula" is more justified.

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 $r$: $$I(r)=\frac{1}{\sqrt{\pi r}}I\left(\frac{1}{\pi^2 r}\right).$$

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.