Integral of $\int_{\mathbb{R}}e^{-\frac{x^{2}}{2}}\left(\cos\left(\pi nx\right)\right)dx$ I was in need to urgently solve this integral. I already know the result in the closed form, does anybody know how to solve it? 
\begin{equation}
 \int_{\mathbb{R}}e^{-\frac{x^{2}}{2}}\left(\cos\left(\pi nx\right)\right)dx=\sqrt{2\pi}e^{-\frac{n^{2}\pi^{2}}{2}},
\end{equation}
If anybody already knows it is welcome, otherwise I guess I will have to look tomorrow with more calm
 A: Here's a funny one:
Define the function $f$ as
$$f:\mathbb{R}\longrightarrow\mathbb{R},\ a\longmapsto\int_{-\infty}^{+\infty}\mathrm{e}^{-x^2/2}\cos(a x)\,\mathrm{d}x.$$
It is well-known that the improper integral defining $f$ is convergent. We can apply the differentiation theorem to differentiate this improper integral (proof left to the reader) and we obtain:
$$\forall a\in\mathbb{R},\ f'(a)=-\int_{-\infty}^{+\infty}x\mathrm{e}^{-x^2/2}\sin(ax)\,\mathrm{d}x.$$
Using an integration by parts yields
$$\forall a\in\mathbb{R},\ f'(a)=-a\int_{-\infty}^{+\infty}\mathrm{e}^{-x^2/2}\cos(a x)\,\mathrm{d}x=-af(a).$$
Hence $f$ is a solution of the differential equation
$$f'(a)+af(a)=0,$$
the general solution of which is
$$f(a)=C\mathrm{e}^{-a^2/2}.$$
Now it is well-known that:
$$f(0)=\int_{-\infty}^{+\infty}\mathrm{e}^{-x^2/2}\,\mathrm{d}x=\sqrt{2\pi},$$
hence
$$\forall a\in\mathbb{R},\ f(a)=\sqrt{2\pi}\mathrm{e}^{-a^2/2}.$$
In particular,
$$\int_{-\infty}^{+\infty}\mathrm{e}^{-x^2/2}\cos(\pi nx)\,\mathrm{d}x=f(\pi n)=\sqrt{2\pi}\mathrm{e}^{-n^2\pi^2/2}.$$
A: Another chance is given by the expansion of $\cos(\pi n x)$ as a Taylor series:
$$ \cos(\pi n x) = \sum_{m\geq 0}\frac{(-1)^m (\pi n)^{2m} x^{2m}}{(2m)!}\tag{1} $$
together with the fact that:
$$ \int_{\mathbb{R}} x^{2m} e^{-x^2/2}\,dx = \int_{0}^{+\infty} x^{m-1/2} e^{-x/2}\,dx = 2^{m+1/2}\cdot\Gamma\left(m-\frac{1}{2}\right).\tag{2} $$
These identities give:
$$\begin{eqnarray*} \int_{\mathbb{R}}e^{-x^2/2}\cos(\pi n x)\,dx &=& \sum_{m\geq 0}\frac{(-1)^m (\pi n)^{2m}\, 2^{m+1/2}\,\Gamma(m+1/2)}{\Gamma(2m+1)} \\ &\color{red}{=}&\sqrt{2\pi}\cdot \sum_{m\geq 0}\frac{(-1)^m (\pi n)^{2m}\, 2^{m}\,\Gamma(m+1/2)}{2^{2m}\,\Gamma(m+1/2)\,\Gamma(m+1)}\\&=&\sqrt{2\pi}\sum_{m\geq 0}\frac{(-1)^m (\pi^2 n^2/2)^m}{m!}\\&\color{blue}{=}&\sqrt{2\pi}\cdot e^{-\pi^2 n^2/2}\tag{3}\end{eqnarray*} $$
where the equality in red follows from the Legendre duplication formula and the equality in blue from the Taylor series of the exponential function.
A: HINT:
$$
\int_{\mathbb{R}}e^{-\frac{x^{2}}{2}}\left(\cos\left(\pi nx\right)\right)dx=
\Re\left[ \int_{\mathbb{R}}e^{-\frac{x^{2}}{2}}e^{i\pi nx}dx\right]
$$
Then observe that
$$
-\frac{x^2}2+i\pi nx=-\left(\frac x{\sqrt2}-\sqrt2i\pi n\right)^2+2\pi^2 n^2
$$
and use the well known integral
$$
\int_{\Bbb R}e^{-s^2}\,ds=\sqrt{\pi}\;\;
$$
with a suitable change of variable.
