Have fun with integral

$\begin{aligned} I &= \int_0^\infty e^{-x^2}{\color{#3D99F6}\cos (2x)} \ dx & \small \color{#3D99F6} \text{By Euler's formula} \\ &= \int_0^\infty e^{-x^2}\cdot {\color{#3D99F6} \frac{e^{2xi}+e^{-2xi}}2} \ dx \\ &= \frac 12 \int_0^\infty \left(e^{(ix+1)^2-1} + e^{(ix-1)^2-1} \right) dx & \small \color{#3D99F6} \text{erfi }(z) = \frac 2 {\sqrt \pi}\int e^{z^2}dz \\ &= \frac {\sqrt \pi}{4ei}(\text{erfi }(ix+1)+ \text{erfi }(ix-1))\bigg|_0^\infty & \small \color{#3D99F6} \text{Note that }\text{erfi }(z) = -i\text {erf }(iz) \\ &= \frac {\sqrt \pi}{4ei}(-i\text{erf }(-x+i) - i \text{erf }(-x-i))\bigg|_0^\infty \\ &= \frac {\sqrt \pi}{4e}(1 + \text{erf }(i) +1 + \text{erf }(-i)) \\ &= \frac {\sqrt \pi}{4e}(2 + i\text{erf }(1) -i \text{erf }(1)) \\ &= \frac {\sqrt \pi}{2e} \approx \boxed{0.326} \end{aligned}$

Relevant wiki: Contour Integration

Since the integrand is even $\int_0^\infty e^{-x^2}\cos 2x\,dx = \frac{1}{2} \int_{-\infty}^\infty e^{-x^2}\cos 2x\,dx$ and since $\cos 2x = \text{Re}[e^{2ix}]$ , $\frac{1}{2} \int_{-\infty}^\infty e^{-x^2}\cos 2x\,dx = \frac{1}{2} \text{Re} \int_{-\infty}^\infty e^{-x^2 + 2ix}\,dx = \frac{1}{2e} \text{Re} \int_{-\infty}^\infty e^{-(x-i)^2}\,dx$

Now, $e^{-z^2}$ is entire and it goes to zero uniformly on $\text{Re}[z] = \pm R, \quad -1 \leq \text{Im}[z] \leq 0$ as $R\to\infty$ , so we can conclude by contour integration around the rectangle with vertices $\pm R, \pm R - i$ that $\int_{-\infty}^\infty e^{-(x-i)^2}\,dx = \int_{-\infty}^\infty e^{-x^2}\,dx$

This last integral is real-valued, so the integral in question equals $\frac{1}{2e} \int_{-\infty}^\infty e^{-x^2}\,dx = \frac{1}{2e} \cdot \sqrt{\pi} \approx \boxed{0.326024666}$

Solution without complex numbers:

A problem like this is clearly meant to be solved by relating it to the Gaussian integral. However, the connection is not obvious so let's try the following standard trick: Let $F(t) = \int_0^{\infty} e^{-x^2} \cos (2tx) dx$ . Then $F'(t) = \int_0^{\infty} e^{-x^2} (-2x \sin(2tx)) dx = \int_0^{\infty} -2xe^{-x^2} \sin(2tx) dx$ . The previous form makes an integration by parts obvious: let $u = \sin (2tx) \implies du = 2t \cos (2tx)$ and $dv = -2xe^{-x^2} \implies v = e^{-x^2}$ . Thus:

$F'(t) = \sin(2tx) e^{-x^2} \Big|_0^{\infty} - \int_0^{\infty} e^{-x^2} 2t \cos(2tx) = 0 - 2t \int_0^{\infty} e^{-x^2} \cos(2tx) = -2t F(t)$ This is now a differential equation that can be solved through standard methods: $F'(t) = -2tF(t) \implies \frac{F'(t)}{F(t)} = -2t \implies \log F(t) = -t^2 + C \implies F(t) = k e^{-t^2}$

We only need to find the constant $k$ but notice that $k = k e^{-0^2} = F(0) = \int_0^{\infty} e^{-x^2} dx = \frac{\sqrt{\pi}}{2}$ . Thus, $F(t) = \frac{\sqrt{\pi}}{2} e^{-t^2}$ . For the problem, we care about $F(1) = \frac{\sqrt{\pi}}{2e}$