Calculating $\lim \limits_{n \to \infty} \int_ {-\infty}^\infty e^{-x^2}\cos(nx)\, dx$ I am trying to calculate
$$\lim \limits_{n \to \infty} \int_ {-\infty}^\infty e^{-x^2}\cos(nx)\, dx$$
Using Fourier transform, but got stuck because of the cosine and the limit involved in the integral. any help will be much appreciated, I will also appreciate if someone could give me some guidelines for calculating limits using Fourier transforms in general...
 A: Hint:
Integration by parts:
$$\int_ {-\infty}^\infty e^{-x^2}\cos(nx)\,\mathrm  dx=\frac 1n \mathrm e^{-x^2}\sin nx\Biggr|_ {-\infty}^\infty+\frac 2n\int_ {-\infty}^\infty x e^{-x^2}\sin(nx)\,\mathrm  dx. $$
A: 
I thought it might be instructuve and of interest to present an approach to evaluating the integral $\int_{-\infty}^\infty e^{-x^2}\cos(nx)\,dx$ that does not rely on direct integration.  To that end we proceed.


Let $f(y)$ be represented by 
$$f(y)=\int_{-\infty}^\infty e^{-x^2}\cos(xy)\,dx \tag1$$
Differentiating $(1)$ under the integral reveals
$$f'(y)=-\int_{-\infty}^\infty xe^{-x^2}\sin(xy)\,dx\tag2$$
Integrating by parts the integral in $(2)$ with $u=-\sin(xy)$ and $v=-\frac12e^{-x^2}$, we obtain
$$\begin{align}
f'(y)&=-\frac12y\int_{-\infty}^\infty e^{-x^2}\cos(xy)\,dx\\\\\
&=-\frac12yf(y)\tag3
\end{align}$$
From $(3)$, we see that $f(y)$ satisfies the ODE $f'(y)+\frac12yf(y)=0$, subject to $f(0)=\sqrt\pi$.  The solution to this ODE is trivial and is given by
$$f(y)=\sqrt\pi e^{-y^2/4}\tag4$$ 
Setting $y=n$ in $(4)$ yields
$$\int_{-\infty}^\infty e^{-x^2}\cos(nx)\,dx=\sqrt\pi e^{-n^2/4}$$
Letting $n\to \infty$, we find the coveted limit is $0$.
A: The integral isn't that difficult to compute, actually, so you could just compute its value and then take the limit.
\begin{align}
\int_{-\infty}^\infty e^{-x^2}\cos(nx)\, dx &= \operatorname{Re}\int_{-\infty}^\infty e^{-x^2 + inx}\,dx
\end{align}
Completing the square in the exponent, we have
\begin{align}
-(x^2-inx) &=-\left(x-\frac{in}{2}\right)^2-\frac{n^2}{4}
\end{align}
Therefore, we have
\begin{align}
\int_{-\infty}^\infty e^{-x^2}\cos(nx)\, dx &= e^{-\frac{n^2}{4}}\operatorname{Re} \int_{-\infty}^\infty\exp\left(-\left(x-\frac{in}{2}\right)^2\right)\,dx
\end{align}
Letting $u = x - \frac{in}{2}$, this becomes a Gaussian integral (after shifting to path of integration back down to the real axes, which can be justified by considering a rectangular contour and realizing that the vertical parts tend to zero as the length of the rectangle becomes infinite) and have
\begin{align}
\lim\limits_{n\rightarrow\infty}\int_{-\infty}^\infty e^{-x^2}\cos(nx)\, dx = \sqrt{\pi} \lim\limits_{n\rightarrow\infty}e^{-\frac{n^2}{4}} = 0
\end{align}
