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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 $7$):

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

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 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 $\alpha=1$.

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. Thus it seems that T. Amdeberhan knew the answer to the question he asked. Why did he ask it then?

I provide a screenshot below for the reader's convenience (integral number $7$):

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

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 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 $\alpha=1$.

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 $7$):

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

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.

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 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 $\alpha=1$.

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 $7$):

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

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 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 $\alpha=1$.

UPDATE

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

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

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.

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 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 $\alpha=1$.

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 $7$):

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

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.

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 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 $\alpha=1$.