Calculating $\lim \limits_{n \to \infty} \int_ {-\infty}^\infty e^{-x^2}\cos(nx)\, dx$

Question

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...

Answer 1

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.

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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 $0$.

Answer 2

Hint:

Integration by parts: $$\int_ {-\infty}^\infty e^{-x^2}\cos(nx)\,\mathrm dx=\frac 1n \mathrm e^{-x^2}\sin nx\Biggr|_ {-\infty}^\infty+\frac 2n\int_ {-\infty}^\infty x e^{-x^2}\sin(nx)\,\mathrm dx. $$

Answer 3

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}