Can somebody evaluate the following Gaussian integral?

$$ I(t,\sigma) := \int_{-\infty}^\infty \frac{dx e^{-x^2/(2\sigma^2)}}{\sqrt{2\pi \sigma^2}} \frac{\sin{\left(2 t\sqrt{1+x^2} \right )}}{\sqrt{1+x^2}} \tag{1} $$

Where of course $\sigma>0$ (and $t\in\mathbb{R}$). Expanding the $\sin$ I could not resum the series. However I have some hope (perhaps misplaced) that the integral can be expressed in terms of elementary functions. In fact an analogous integral (that arises in the same problem) turns out to be surprisingly simple:

\begin{align} G(t,\sigma) &:= \int_{-\infty}^\infty \frac{dx e^{-x^2/(2\sigma^2)}}{\sqrt{2\pi \sigma^2}} \tag{2} \frac{1+x^2\cos{\left(2 t\sqrt{1+x^2} \right )}}{1+x^2}\\ &\simeq \exp\left( -2 \sigma^2 \sin(t )^2 \right) \end{align}


My memory failed me in a more complicated way. It turns out that the above equation is not an exact evaluation of the integral $G$ but instead an excellent approximation (based on some physical theory).

To my excuse here is a plot of $G(t,\sigma=0.2)$ (dots) versus the approximation (continuous line)

enter image description here

At this point I have little hope that $I$ (or $G$) can be still evaluated analytically but I would love to be proved wrong.

Thanks again!

  • 1
    $\begingroup$ Can you share the derivation for the second integral G? $\endgroup$ May 13 '20 at 12:54
  • $\begingroup$ @thomasfermi it's a bit embarrassing but I completely forgot. I found the expression in my notes. I checked numerically and appears to be right. But I probably used some brute force method, whereas maybe a smart substitution could be applied to the other integral too. $\endgroup$
    – lcv
    May 13 '20 at 17:34

I managed to get this into a series, but not a closed form. I got that $$I = \sqrt{t\pi}\sum_{k=0}^{\infty}\left(-\frac{\sigma^2t}{2}\right)^k \binom{2k}{k}J_{1/2 + k} (2t)$$

The question is to solve

$$\int_{-\infty}^\infty \frac{e^{-x^2/(2\sigma^2)}}{\sqrt{2\pi \sigma^2}} \frac{\sin{\left(2 t\sqrt{1+x^2} \right )}}{\sqrt{1+x^2}} dx$$

The constant $\frac{1}{\sqrt{2\pi\sigma^2}}$ can be taken out of the integral to get

$$\frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^\infty e^{-x^2/(2\sigma^2)} \frac{\sin{\left(2 t\sqrt{1+x^2} \right )}}{\sqrt{1+x^2}} dx$$

Here the $\sin$ can be expanded into a sum to get

$$\frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^\infty e^{-x^2/(2\sigma^2)} \frac{\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}\left(2 t\sqrt{1+x^2} \right )^{2n+1}}{\sqrt{1+x^2}} dx$$

Swapping the sum and the integral, I get

$$\frac{1}{\sqrt{2\pi\sigma^2}}\sum_{n=0}^{\infty}\int_{-\infty}^\infty e^{-x^2/(2\sigma^2)} \frac{\frac{(-1)^n}{(2n+1)!}\left(2 t\sqrt{1+x^2} \right )^{2n+1}}{\sqrt{1+x^2}} dx$$

The constants with respect to $x$ can be taken out of the integral

$$\frac{1}{\sqrt{2\pi\sigma^2}}\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} (2t)^{2n+1}\int_{-\infty}^\infty e^{-x^2/(2\sigma^2)} (1+x^2)^n dx$$

Here $(1+x^2)^n$ can be expanded to get

$$\frac{1}{\sqrt{2\pi\sigma^2}}\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} (2t)^{2n+1}\int_{-\infty}^\infty e^{-x^2/(2\sigma^2)} \sum_{k=0}^n \binom{n}{k} x^{2k} dx$$

The sum inside the integral can be taken out to get $$\frac{1}{\sqrt{2\pi\sigma^2}}\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} (2t)^{2n+1}\sum_{k=0}^n \binom{n}{k}\int_{-\infty}^\infty e^{-x^2/(2\sigma^2)} x^{2k} dx$$

The inner integral can be calculated in closed form, so this simplifies to

$$\frac{1}{\sqrt{2\pi\sigma^2}}\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} (2t)^{2n+1}\sum_{k=0}^n \binom{n}{k}\sigma^{1 + 2k} \frac{(2k)!}{2^kk!}\sqrt{2\pi}$$

This reduces to $$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} (2t)^{2n+1}\sum_{k=0}^n \binom{n}{k}\sigma^{2k} \frac{(2k)!}{2^kk!} \tag 1$$

Changing the order of summation from $n, k$ to $k, n$, I get $$\sum_{k=0}^{\infty}\sigma^{2k} \frac{(2k)!}{2^kk!}\sum_{n=k}^\infty \frac{(-1)^n}{(2n+1)!} (2t)^{2n+1} \binom{n}{k}$$

Mathematica tells me that the inner sum can be written as $\frac{(-1)^k \sqrt{\pi} t^{1/2 + k} J_{1/2 + k} (2t)}{k!}$ where $J$ is the Bessel function of the first kind. This leads to $$\sqrt{t\pi}\sum_{k=0}^{\infty}\left(-\frac{\sigma^2t}{2}\right)^k \binom{2k}{k}J_{1/2 + k} (2t)$$

  • $\begingroup$ Thanks! (+1). Unfortunately this seems to be a worse approximation than the simple Taylor series (1) (obtained without exchanging summations). Probably because there you get twice as many powers of $t$ and the Bessel function does not decay fast enough. But this is indeed a more explicit formula. $\endgroup$
    – lcv
    May 14 '20 at 21:12
  • 1
    $\begingroup$ Probably useless, but I got this alternative representation: $$I(t,\sigma)= t\int_0^1 {e^{ - (\sigma t)^2 x} I_0 \left( { - (\sigma t)^2 x} \right)\frac{{\cos (2t\sqrt {1 - x} )}}{{\sqrt {1 - x} }}dx} . $$ The $I_0$ denotes the modified Bessel function. I derived it from VVejalla's series by replacing the Bessel functions with one of their integral represnetations (dlmf.nist.gov/10.9.E4), changed the order of summation and integration, and used the exponential generating function of the central binomials. $\endgroup$
    – Gary
    May 15 '20 at 9:11
  • $\begingroup$ @lcv If you like, you could write the inner sum in $(1)$ as a hypergeometric function, but I don't think it will be much help. $\endgroup$ May 15 '20 at 19:45
  • 1
    $\begingroup$ @Gary Neat! Maybe with substitution by parts using your representation, something could be done? $\endgroup$ May 15 '20 at 19:46
  • $\begingroup$ @Gary Thanks that looks non-trivial. Pity one cannot get anything out of it. I'll give it a look. (Btw if you want to post it as an answer you will get at least one upvote) $\endgroup$
    – lcv
    May 16 '20 at 1:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.