Just another gaussian problem

Ok, both solutions are very interesting. They are both nice and imaginative. Mine is a little less imaginative:

$\int\limits_{-\infty }^{+\infty }{\cos (x){{e}^{-\frac{{{x}^{2}}}{2}}}dx}=\int\limits_{-\infty }^{+\infty }{(\cos (x)+i\sin (x)){{e}^{-\frac{{{x}^{2}}}{2}}}dx}=\int\limits_{-\infty }^{+\infty }{{{e}^{ix}}{{e}^{-\frac{{{x}^{2}}}{2}}}dx}=$ $\int\limits_{-\infty }^{+\infty }{{{e}^{-\frac{{{x}^{2}}}{2}+ix}}dx}=\int\limits_{-\infty }^{+\infty }{{{e}^{-\frac{{{x}^{2}}-2ix}{2}}}dx}=\int\limits_{-\infty }^{+\infty }{{{e}^{-\frac{{{x}^{2}}-2ix+{{i}^{2}}}{2}}}{{e}^{\frac{{{i}^{2}}}{2}}}dx}=$ ${{e}^{-\frac{1}{2}}}\int\limits_{-\infty }^{+\infty }{{{e}^{-\frac{{{(x-i)}^{2}}}{2}}}dx}={{e}^{-\frac{1}{2}}}\sqrt{2\pi }=\sqrt{\frac{2\pi }{e}}$

Using the gaussian result. Like both the other solutions, it's lacks somehow of a mathematical prove (the exchange between sum and integral for infinite limits of integration and the possibility of derivative under an integral with infinite range). The gaussian integral is not a shift on the real numbers. The prove is easy, though! Thanks to all. I've learned a lot!

Let $I(\alpha)=\int_{-\infty }^{\infty} cos(\alpha x)e^{-x^2/2}dx$ $\frac{\mathrm{d} I }{\mathrm{d} \alpha}=-\int_{-\infty }^{\infty} xsin(\alpha x)e^{-x^2/2}dx$

using integrating by parts : $\frac{\mathrm{d} I }{\mathrm{d} \alpha} = \left [ sin(\alpha x)e^{-x^2/2} \right ]_{-\infty}^{\infty} -\alpha\int_{-\infty}^{\infty}e^{-x^2/2}cos(\alpha x) dx$ $\frac{\mathrm{d} I }{\mathrm{d} \alpha} = 0-\alpha I$ therefore, $I(\alpha)=(const)e^{-\alpha^2/2}$ we know that $I(0)=\int_{-\infty }^{\infty}e^{-x^2/2}dx=\sqrt{2\pi}$ $const=\sqrt{2\pi}$ thus, $\int_{-\infty }^{\infty} cos(x)e^{-x^2/2}dx = \sqrt{2\pi/e} =1.52$

Let start with infinite series!

Fisrt we have $\displaystyle\cos(x)=\Sigma_{n=0}^{\infty} \frac{x^{2n}(-1)^{n}}{(2n)!}$

So the integral becomes

$\displaystyle \int_{-\infty}^{\infty} \cos(x)e^{-\frac{x^{2}}{2}}dx=\int_{-\infty}^{\infty} (\Sigma_{n=0}^{\infty} \frac{x^{2n}(-1)^{n}}{(2n)!})e^{-\frac{x^{2}}{2}}dx$

$\displaystyle \Sigma_{n=0}^{\infty}(\int_{-\infty}^{\infty} \frac{x^{2n}(-1)^{n}}{(2n)!}e^{-\frac{x^{2}}{2}}dx)$

$\displaystyle \int_{-\infty}^{\infty} \frac{x^{2n}(-1)^{n}}{(2n)!}e^{-\frac{x^{2}}{2}}dx=\frac{(-1)^{n}}{(2n)!}\times (2n-1) \int_{-\infty}^{\infty} x^{2n-3}e^{-\frac{x^{2}}{2}}dx$

Then $\displaystyle \int_{-\infty}^{\infty} x^{2n}e^{-\frac{x^{2}}{2}}dx=\frac{(2n)!}{2^{n}n!} \int_{-\infty}^{\infty} e^{-\frac{x^{2}}{2}}dx$

$\displaystyle \Sigma_{n=0}^{\infty}(\int_{-\infty}^{\infty} \frac{x^{2n}(-1)^{n}}{(2n)!}e^{-\frac{x^{2}}{2}}dx) = \Sigma_{n=0}^{\infty} \frac{(-1)^{n}}{2^{n}n!}\times \int_{-\infty}^{\infty} e^{-\frac{x^{2}}{2}}dx$

$\displaystyle \Sigma_{n=0}^{\infty} \frac{(-1)^{n}}{2^{n}n!} =e^{-\frac{1}{2}}$

so the answer is

$\displaystyle e^{-\frac{1}{2}}\times \sqrt{2\pi}=\sqrt{\frac{2\pi}{e}}$

Similar solution with @Humberto Bento 's

$\begin{aligned} I & = \int_{-\infty}^\infty \cos x \ e^{-\frac {x^2}2} dx & \small \color{#3D99F6} \text{By Euler's formula: }e^{i\theta} = \cos \theta + i \sin \theta \\ & = \int_{-\infty}^\infty \Re \left(e^{ix} e^{-\frac {x^2}2}\right) dx & \small \color{#3D99F6} \text{where }\Re (z) \text{ is the real part of }z. \\ & = \Re \left(\int_{-\infty}^\infty e^{-\frac 12 (x^2-2ix)} dx \right) \\ & = \Re \left(\int_{-\infty}^\infty e^{-\frac 12 ((x-i)^2+1)} dx \right) & \small \color{#3D99F6} \text{Let }u = \frac {x-i}{\sqrt 2} \implies du = \frac {dx}{\sqrt 2} \\ & = \Re \left(\sqrt {\frac 2e} \color{#3D99F6} \int_{-\infty}^\infty e^{-u^2} du \right) & \small \color{#3D99F6} \text{Note that } \int_{-\infty}^\infty e^{-u^2} du \text{ is Gaussian integral.} \\ & = \Re \left(\sqrt {\frac 2e} \color{#3D99F6} \sqrt \pi \right) \\ & = \sqrt {\frac {2\pi}e} \approx \boxed{1.520} \end{aligned}$

integral_(-infinity)^infinity (cos(x))/e^(x^2/2) dx = sqrt((2 pi)/e)~~1.5203469010662808056119401467549756270361074187790463375283638685267346239300583043148415372595655771