**UPDATE**

The integral in T. Amdeberhan's question was taken from his own paper that was written 10 years earlier before he posted his question: https://arxiv.org/abs/0808.2692 *A dozen integrals: Russell-style*. I provide a screenshot below for reader's convenience (integral number $7$):

[![enter image description here][1]][1]

At the end of the paper, the authors provide a sketch of proof:

[![enter image description here][2]][2]

Thus T. Amdeberhan knew the answer to the question he asked. Why did he ask it then??? I think he asks questions taken from well known textbooks, folklore results, forums including math stack exchange, and posts them on MO to farm reputation.

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**OLD ANSWER**

Note that the following functions are self-reciprocal $\sqrt{\frac{2}{\pi}}\int_0^\infty f(x)\cos ax dx=f(a)$ (see this [MO post][3] for a list of such functions):
$$
\frac{1}{\cosh\sqrt{\frac{\pi}{2}}x},\quad e^{-x^2/2}.\tag{1}
$$
It was proved by Hardy (Quarterly Journal Of Pure And Applied Mathematics, Volume 35, Page 203, 1903) and Ramanujan (Ramanujan's Lost Notebook, part IV, chapter 18) that for two self-reciprocal functions $f$ and $g$ we have
$$
\int_0^\infty f(x)g(\alpha x) dx=\frac{1}{\alpha}\int_0^\infty f(x)g(x/\alpha) dx.
$$
The formal agument is as follows but it can be made rigorous for certain type of functions
$$
\int_0^\infty f(x)g(\alpha x) dx=\int_0^\infty f(x)dx\cdot \sqrt{\frac{2}{\pi}}\int_0^\infty g(y) \cos(\alpha x y)dy=\\
\int_0^\infty g(y)dy \cdot \sqrt{\frac{2}{\pi}}\int_0^\infty f(x)\cos(\alpha x y)dx=\int_0^\infty g(y)f(\alpha y) dy=\\
\frac{1}{\alpha}\int_0^\infty f(x)g(x/\alpha) dx.
$$
For functions in $(1)$ which decay rapidly at $x\to\pm\infty$ it is certainly true. So we obtain an identity due to Hardy and Ramanujan
$$
\int_{0}^{\infty} \frac{e^{-x^{2}/2}}{\cosh{\sqrt{\frac{\pi}{2}}\alpha{x}}} {dx} = \frac{1}{\alpha} \int_{0}^{\infty} \frac{e^{-x^{2}/2}}{\cosh{\sqrt{\frac{\pi}{2}}\frac{x}{\alpha}}}{dx}.
$$
After some simplifications it becomes 
$$
\int_{0}^{\infty} \frac{e^{-\alpha^2x^{2}}}{\cosh{\sqrt{{\pi}}{x}}} {dx} = \frac{1}{\alpha} \int_{0}^{\infty} \frac{e^{-x^{2}/\alpha^2}}{\cosh{\sqrt{{\pi}}{x}}}{dx}.
$$
To complete the proof differentiate this with respect to $\alpha$ and then set $\alpha=1$.


  [1]: https://i.sstatic.net/sxjSB.png
  [2]: https://i.sstatic.net/NeNcP.png
  [3]: https://mathoverflow.net/a/224201/82588