I have tried to calculate the following integral in terms of $\sigma^2$ :

$I=\int_{x=-\infty}^{+\infty}\frac{1}{\sqrt{1+x^2}}e^{-x^2/{2\sigma^2}} dx$

However by using conventional methods such as changing variables I canmot reach to a closed-form solution even with Q-function.

I appreciate any help in determining the solution of this integral.

My unsuccessfull tries are:

1- change of variable $u^2 := \frac{1}{\sqrt{1+x^2}}$ is resulted a nother more complicated gaussian Integral of this form.

2-representing in the polar system also have the same problem as above.

3-using enter link description here does not give answer.

4-checking enter link description here

5-integral by parts.

  • 1
    $\begingroup$ Welcome to Mathematics Stack Exchange! As this site is meant to be a useful repository rather than a Do My Homework forum, it's common courtesy to show what you've already tried, and really narrow down what you're struggling with. Most people here are glad to help once you've adequately motivated the problem. Quick Guide to attracting answers and preventing your question from being deleted. Good luck! $\endgroup$
    – linkja
    Nov 21, 2023 at 15:12
  • $\begingroup$ @linkja Is it better? $\endgroup$ Nov 21, 2023 at 15:52
  • 1
    $\begingroup$ $$I=2\int_0^\infty\frac{e^{-x^2/{2\sigma^2}} }{\sqrt{1+x^2}}dx\overset{x=\sinh t}{=}2\int_0^\infty e^{-\frac{\sinh^2t}{2\sigma^2}}dt=2e^{\frac1{4\sigma^2}}\int_0^\infty e^{-\frac{\cosh 2t}{4\sigma^2}}dt=e^{\frac1{4\sigma^2}}K_0\Big(\frac1{4\sigma^2}\Big)$$ dlmf.nist.gov/10.32#E9 $\endgroup$
    – Svyatoslav
    Nov 22, 2023 at 0:17


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