Gaussian integral, but with a hyperbolic cosine.

$\begin{aligned} I & = \int_0^\infty e^{-x^2} \cosh x \ dx \\ & = \int_0^\infty e^{-x^2} \left(\frac {e^x+e^{-x}}2\right) dx \\ & = \int_0^\infty \frac {e^{-x^2+x}+e^{-x^2-x}}2 dx \\ & = \int_0^\infty \frac {e^{-\left(x-\frac 12\right)^2 + \frac 14}+e^{-\left(x+\frac 12\right)^2 + \frac 14}}2 dx & \small \color{#3D99F6} \text{Let }u = x - \frac 12,\ v = x + \frac 12 \\ & = \frac {\sqrt[4] e}2 \left(\int_{-\frac 12}^\infty e^{-u^2} du + \int_\frac 12^\infty e^{-v^2} dv \right) & \small \color{#3D99F6} \text{Replace }u \text{ and }v \text{ with }t. \\ & = \frac {\sqrt[4] e}2 \left({\color{#3D99F6}\int_{-\frac 12}^0 e^{-t^2} dt} + \int_0^\infty e^{-t^2} dt + \int_0^\infty e^{-t^2} dt - \int_0^\frac 12 e^{-t^2} dt \right) & \small \color{#3D99F6} \text{Note that } \int_{-\frac 12}^0 e^{-t^2} dt \text{ is an even function.} \\ & = \frac {\sqrt[4] e}2 \left({\color{#3D99F6}\int_0^{\frac 12} e^{-t^2} dt} + 2 \int_0^\infty e^{-t^2} dt - \int_0^\frac 12 e^{-t^2} dt \right) \\ & = \sqrt[4] e \int_0^\infty e^{-t^2} dt & \small \color{#3D99F6} \text{Let }s = t^2 \implies ds = 2t \ dt \\ & = \frac {\sqrt[4] e}2 \int_0^\infty s^{-\frac 12} e^{-s} ds & \small \color{#3D99F6} \text{Gamma function }\Gamma (z) = \int_0^\infty t^{z-1} e^{-t} dt \\ & = \frac {\sqrt[4] e}2 \cdot {\color{#3D99F6} \Gamma \left(\frac 12\right)} = \frac {\sqrt[4] e}2 \cdot {\color{#3D99F6} \sqrt \pi} = \boxed{\dfrac {\sqrt{\pi \sqrt e}}2} \end{aligned}$

Let us begin with $\int_{0}^{\infty} e^{-x^2} \cdot cosh(x) dx = \int_{0}^{\infty} e^{-x^2} \cdot (\frac{e^x + e^{-x}}{2}) dx$ ;

or $\frac{1}{2} \cdot \int_{0}^{\infty} e^{-(x^2 - x)} + e^{-(x^2 + x)} dx;$

or $\frac{1}{2} \cdot \int_{0}^{\infty} e^{-(x - \frac{1}{2})^2 + \frac{1}{4}} + e^{-(x + \frac{1}{2})^2 + \frac{1}{4}} dx$ ;

or $\frac{1}{2} \cdot e^{\frac{1}{4}} \cdot \int_{0}^{\infty} e^{-(x - \frac{1}{2})^2} + e^{-(x + \frac{1}{2})^2} dx;$

Now, let $u = x - \frac{1}{2}, du = dx, -\frac{1}{2} \le u < \infty$ AND $v = x + \frac{1}{2}, dv = dx, \frac{1}{2} \le v < \infty.$ This now results in:

$\frac{1}{2} \cdot e^{\frac{1}{4}} \cdot [ \int_{-\frac{1}{2}}^{\infty} e^{-u^2} du + \int_{\frac{1}{2}}^{\infty}e^{-v^2} dv];$

or $\frac{1}{2} \cdot e^{\frac{1}{4}} \cdot [ \int_{-\frac{1}{2}}^{0} e^{-u^2} du + \int_{0}^{\infty} e^{-u^2} du + \int_{0}^{\infty}e^{-v^2} dv - \int_{0}^{\frac{1}{2}} e^{-v^2} dv];$

By exploiting symmetry, we have $\int_{-\frac{1}{2}}^{0} e^{-u^2} du = \int_{0}^{\frac{1}{2}} e^{-v^2} dv$ . This helps simplify our integration into:

$\frac{1}{2} \cdot e^{\frac{1}{4}} \cdot [ \int_{-\frac{1}{2}}^{0} e^{-u^2} du + \int_{0}^{\infty} e^{-u^2} du + \int_{0}^{\infty}e^{-v^2} dv - \int_{0}^{\frac{1}{2}} e^{-v^2} dv] = \frac{1}{2} \cdot e^{\frac{1}{4}} \cdot 2\int_{0}^{\infty} e^{-u^2} du$ ;

or $e^{\frac{1}{4}} \cdot \frac{\sqrt{\pi}}{2};$

or $\boxed{ \frac{\sqrt{\pi \cdot \sqrt{e}}}{2}}.$

We can use the following property: $f \text{ is continuous on } (-a,a) \implies \int_0^a f_{\text{even}}(x)\,dx = \frac{1}{2} \int_{-a}^a f(x)\,dx$ where $f_{\text{even}}$ is the even part of $f$ , defined by $f_{\text{even}}(x) = \frac{f(x) + f(-x)}{2}.$ For instance, $\cosh x$ is defined as the even part of $e^x,$ and since $e^{-x^2}$ is even, we can conclude that $e^{-x^2}\cosh x$ is the even part of $\large e^{-x^2}\cdot e^x = e^{-x^2 + x} = e^{-\left(x-\frac{1}{2}\right)^2 + \frac{1}{4}} = \sqrt[4]{e} \cdot e^{-\left(x-\frac{1}{2}\right)^2}$

Then, $\begin{aligned} \int_0^\infty e^{-x^2}\cosh x\,dx &= \lim_{N\to\infty} \int_0^N e^{-x^2}\cosh x\, dx \\ &= \frac{1}{2} \lim_{N\to\infty} \int_{-N}^N \sqrt[4]{e} \cdot e^{-\left(x-\frac{1}{2}\right)^2}\,dx & \left(\text{By the above property}\right) \\ &= \frac{\sqrt[4]{e}}{2} \lim_{N\to\infty} \int_{-N-\frac{1}{2}}^{N-\frac{1}{2}} e^{-u^2}\,du & \left(\text{Using the substitution } u = x-\frac{1}{2}\right) \\ &= \frac{\sqrt[4]{e}}{2} \int_{-\infty}^{\infty} e^{-u^2}\,du \\ &= \frac{\sqrt[4]{e}}{2} \cdot \sqrt{\pi} \\ &= \boxed{\dfrac{\sqrt{\pi\sqrt{e}}}{2}} \end{aligned}$