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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...

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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 $0$.

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Hint:

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

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  • $\begingroup$ This just makes the integral more difficult, so what ? $\endgroup$ – Yves Daoust Jul 2 '18 at 15:19
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    $\begingroup$ @YvesDaoust I think the key here is that the latter integral is bounded by $\frac{2}{n}\int_0^\infty 2xe^{-x^2}{\rm d}x$ and you can compute this quite easily. $\endgroup$ – Winther Jul 2 '18 at 15:22
  • $\begingroup$ @Winther: indeed. $\endgroup$ – Yves Daoust Jul 2 '18 at 15:25
  • $\begingroup$ Exactly. Anyway, 'tis only a hint, not a full solution. $\endgroup$ – Bernard Jul 2 '18 at 15:29
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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}

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  • $\begingroup$ I don't like the sentence "letting $u=x-\frac{in}{2}$" because it suggests that what you are doing here is a change of variable, while, as you explained just after, the point is to use complex contour integration. I'm pointing out this not to be pedantic, but because indeed it can be very confusing: even my probability teacher and his assistant thought that "it is just the calculus change of variable formula, didn't you pass calculus?". I did, but after that I thought they didn't. $\endgroup$ – Bob Jul 14 '18 at 4:15

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