# Proof for Gaussian integrals of form $\int_{-\infty}^{\infty} {\alpha\over\sqrt{\pi}} e^{-\alpha^2 x^2} f(x) dx$?

I have two equations in question here, and I'm looking for a method of solution that doesn't involve the error function $\mathrm {erf}(z)$, and a proof that the answers given are correct. Both these equations are found in B.E. Warren's X-Ray Diffraction on page 36, if anyone's curious, but I'll outline the necessary context. These two equations are examples of time averages of functions of $x$ for which $x$ follows a Gaussian distribution, the Gaussian probability being expressed by $$p(x) = {\alpha\over\sqrt\pi}e^{-\alpha^2x^2},$$ where $$\int_{-\infty}^{\infty}p(x)\space dx = 1.$$ In the case of the first equation, the term $\langle e^{ix} \rangle$ has been changed to $\langle\cos x\rangle + i\langle\sin x\rangle$, and subsequently just $\langle\cos x\rangle$ considering the average of $\sin x$ for an $x$ which follows Gaussian distribution is equal to zero (or so I'm told) and can consequently be removed. The purpose of these equations is to show that the time average of $e^{ix}$ can be written as $\langle e^{ix} \rangle = e^{-(1/2)\langle x^2\rangle}$.

The equations are as follows, $$\langle e^{ix}\rangle = \int_{-\infty}^{\infty}{\alpha\over\sqrt{\pi}} e^{-\alpha^2 x^2}\cos x \space dx = e^{-(1/4\alpha^2)},$$ $$\langle x^2\rangle = \int_{-\infty}^{\infty}{\alpha\over\sqrt{\pi}} e^{-\alpha^2 x^2}x^2 \space dx = {1\over{2\alpha^2}}.$$ You'll notice that solutions are given. Again, I'm not looking for the solutions, but the method of solutions of these integrals, and proofs that the solutions are as shown. I'll post any more necessary information as needed. Much thanks in advance.

: I'm also curious about Warren's setup of the first integral with $\cos x$ as part of the integrand instead of $e^{ix}$. Is there a reason for this? Most solutions I've gotten to this problem, here and elsewhere, involve the complex exponential instead of the cosine. If that's so, why change it in the first place?

I'm also wondering if the second equation has some Fourier integral-related solution that would remove the need to put it into polar coordinates. Thanks very much.

For the first one observes that $e^{-\alpha^2 x^2}\sin x$ is an odd function, so that its integral over $(-\infty,\infty)$ vanishes. Then $$\int_{-\infty}^{\infty}e^{-\alpha^2 x^2}e^{ix}\,dx=\int_{-\infty}^{\infty} e^{-(\alpha^2 x^2-ix)}\,dx=\int_{-\infty}^{\infty}e^{-(\alpha x-i/(2\alpha))^2-1/(4\alpha^2)}\,dx= e^{-1/(4\alpha^2)}(1/\alpha)\int_{-\infty}^{\infty}e^{-u^2}\, du.$$

For the second one you can use the indication given in user2741736's answer or mimic the standard computation of the Gaussian integral. A change of variables shows that it is enough to consider the case \alpha=1. Let $I=\int_{-\infty}^\infty e^{-\alpha ^2 x^2}x^2\,dx$. Then $$I^2=\int_{-\infty}^\infty e^{-\alpha^2x^2}x^2\,dx\,\int_{-\infty}^\infty e^{-\alpha^2y^2}y^2\,dy=\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-\alpha^2(x^2+y^2)}x^2y^2\,dxdy.$$ Change to polar coordinates to obtain $$I^2=\int_0^\infty\int_0^{2\pi}e^{-\alpha^2r^2}r^5\cos^2\theta\sin^2\theta\,drd\theta.$$ This last integral can be computed by standard techniques.

• Thanks for the detailed response, I'm wondering though how you justify $e^{-(\alpha x+i/(2\alpha))^2}$ becoming $e^{-\alpha^2 x^2}$ with the removal of $e^{-1/(4\alpha^2)}$ because to me it seems that what would remain as the integrand would be $e^{-(\alpha^2 x^2 - ix - 1/(4\alpha^2))}$. – Tidewater Sep 6 '13 at 11:22
• @Tidewater let $\alpha x + i/2\alpha = y$ and get another Gaussian integral. – Santosh Linkha Sep 6 '13 at 11:43

$\frac { \partial\exp(-b \cdot x^2 ) }{ \partial b} = -x^2 \cdot {\exp(-x^2 \cdot b)}$
$\int_{-\infty}^{\infty} \frac { \partial }{ \partial b} \exp(-b \cdot x^2 ) dx = \frac{ \text{d}} {\text{d}b}\int_{-\infty}^{\infty} \exp(-b \cdot x^2)$ dx
• I'm not sure your meaning by that, $\alpha$ here isn't a variable, it's a sort of normalizing constant made to satisfy the integration of the Gaussian probability. The integrals in question are in terms of $x$ and aren't partial. – Tidewater Sep 6 '13 at 9:35