Hello there I am trying to solve for $a > b$: $$I=\int_0^{\infty} \frac{e^{-x^2}}{a+b\cos{x}}dx$$ My thought was to expand into fourier series $$g(t)=\frac{1}{a+b\cos t}$$ Since g(t) has the period $T=2\pi\, $ we can rewrite $$g(t)=\frac{a_0}{2}+\sum_{n=1}^{\infty}(a_n\cos{nt}+b_n\sin{nt})$$ $$a_0=\frac{1}{\pi}\int_0^{2\pi}\frac{1}{a+b\cos t}dt$$ $$a_n=\frac{1}{\pi}\int_0^{2\pi}\frac{\cos(nt)}{a+b\cos t}dt$$$$b_n=\frac{1}{\pi}\int_0^{2\pi}\frac{\sin(nt)}{a+b\cos t}dt$$ $$X=a_n+ib_n=\frac{1}{\pi}\int_0^{2\pi}\frac{e^{int} }{a+b\cos t}dt$$ we let $$e^{it}=z \rightarrow dt=\frac{dz}{iz}\, ; |z|=1$$ And we can write due to Euler's formula $\cos t=\frac{z^2+1}{2z}$ $$X=\frac{2}{i\pi}\oint_{|z|=1} \frac{z^n}{bz^2+2az+b}dz$$ Inside the circle $|z|=1\,$ the integrand function has only the pole $z_1=\frac{\sqrt{a^2-b^2}-a}{b}\,$ Thus our residue is $$\text{Res}(f,z_1)=\lim_{z\to z_1} \frac{z^n}{2bz+2a}=\frac{1}{2}\frac{1}{\sqrt{a^2-b^2}}\left(\frac{\sqrt{a^2-b^2}-a}{b}\right)^n$$ $$X=\frac{2}{\sqrt{a^2-b^2}}\left(\frac{\sqrt{a^2-b^2}-a}{b}\right)^n$$ Thus $a_n=X \, , \, b_n=0$ and finally $$g(t)=\frac{1}{\sqrt{a^2-b^2}}+\frac{2}{\sqrt{a^2-b^2}}\sum_{n=1}^{\infty}\left(\frac{\sqrt{a^2-b^2}-a}{b}\right)^n\cos{nt}$$ Plugging those into the original integral and making use of the gaussian integral for example see here: $\int_{-\infty}^{+\infty} e^{-x^2} dx$ with complex analysis also $$\int_{0}^{\infty} e^{-x^2}\cos(nx)dx=\Re\frac{1}{2}\int_{-\infty}^{\infty} e^{-x^2+nix}dx= \Re\frac{1}{2}\int_{-\infty}^{\infty} e^{-(x-\frac{ni}{2})^2-\frac{n^2}{4}}dx=\frac{\sqrt{\pi}}{2}e^{\frac{-n^2}{4}}$$ and denoting $w=\frac{\sqrt{a^2-b^2}-a}{b}$ will yield to: $$I=\frac{\sqrt{\pi}}{\sqrt{a^2-b^2}}\left(\frac{1}{2}+\sum_{n=1}^{\infty}w^ne^{-\frac{n^2}{4}}\right)$$ Now I never encountered the latter sum, is there already a known closed form or can you help me find one? Also have I done any mistakes?

  • $\begingroup$ The integral doesn't even converge in general unless $a>b$. $\endgroup$ – Mark Viola Jun 15 '18 at 18:38
  • $\begingroup$ For starters, it seems like that integral does not necessarily converge as $a + b \cos x$ can equal 0 $\endgroup$ – Gregory Jun 15 '18 at 18:38
  • 2
    $\begingroup$ That sum is a Theta function: en.wikipedia.org/wiki/Theta_function $\endgroup$ – marty cohen Jun 15 '18 at 18:54
  • 2
    $\begingroup$ @martycohen. That sum looks like Theta function,but it is not. $\endgroup$ – Mariusz Iwaniuk Jun 15 '18 at 21:10
  • 1
    $\begingroup$ I'm pretty sure that there is no closed from solution for this integral :( $\endgroup$ – tired Jun 16 '18 at 8:47

Your Answer

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

Browse other questions tagged or ask your own question.