Is it possible to evaluate this integral in closed form?

$$I(a)=\int_0^\infty \frac{e^{-x^2}}{1+a \cos x}dx$$


I encountered this integral when trying to find a closed form for the series from this question.

Denominator doesn't have any zeroes on the real line, however it does have complex zeroes. I'm not sure how to use residues here, because the usual methods involve either a polynomial in the denominator with infinite limits, or trigonometric functions but with limits $\pm \pi/2$. Which is not the case here.

Another way would be to expand the denominator as the Taylor series, but I don't know the closed form for $\int_0^\infty e^{-x^2} \cos^k x dx$ much less the resulting series.

The function under the integral is even, so the limits can be extended to $\pm \infty$.

  • 1
    $\begingroup$ $$\int_0^\infty e^{-x^2}\cos nx$$ is a value of the Fourier transform of $e^{-x^2}$ which is well-known. You can express $\cos^k x$ as a linear combination of the $\cos nx$. $\endgroup$ – Lord Shark the Unknown Apr 11 '18 at 8:20
  • $\begingroup$ @LordSharktheUnknown, this is exactly what I used to obtain the integral in question. So it wouldn't make any sense to go back $\endgroup$ – Yuriy S Apr 11 '18 at 8:21
  • $\begingroup$ The original question is the sort of question that the Poisson summation formula handles pretty well. $\endgroup$ – Jack D'Aurizio Apr 11 '18 at 19:11

$\int_0^{+\infty}\frac{e^{-x^2}}{1+a\cos x}~dx=\frac12\sqrt{\frac\pi{1-a^2}}\left(1+2\sum_{n=1}^\infty (-1)^n e^{-\frac{n^2}4}\alpha^n\right)$ where $\alpha=\frac1a-\sqrt{\frac1a^2-1}$

First I intoduce the auxiliary integral:$$\int_0^{2\pi}\frac{\cos(nx)}{1+a \cos(x)}~dx=\frac{2\pi(-1)^n \alpha ^n}{\sqrt{1-a^2}}$$ where $\alpha=\frac1a-\sqrt{\frac1a^2-1}$ and $n\in\mathbb{Z}$.

Proof. $\triangleright$ $$ \begin{align} \int_0^{2\pi}\frac{\cos(nx)}{1+a \cos(x)}~dx&=\Re\int_0^{2\pi}\frac{e^{nix}}{1+a \cos(x)}~dx=\left[ \begin{matrix} z=e^{ix}\\ \cos x=\frac12\left(z+\frac1z\right)\\ dx=\frac{dz}{iz} \end{matrix} \right]\\ &=\Re\int_{|z|=1}\frac{z^n}{1+\frac a2\left(z+\frac1z\right)}\frac{dz}{iz} =\Re\int_{|z|=1}\frac{-2iz^n}{az^2+2z+a}~dz\\ &=\Re\left(2\pi i\operatorname*{Res}_{z=-\frac1a+\sqrt{\frac1a^2-1}}\frac{-2iz^n}{az^2+2z+a}\right)=\frac{2\pi(-1)^n \alpha ^n}{\sqrt{1-a^2}} \end{align}$$ where $\alpha=\frac1a-\sqrt{\frac1a^2-1}$ $\triangleleft$.

Finally let's crack the initial integral

$$\begin{align} I=\int_0^{+\infty}\frac{e^{-x^2}}{1+a\cos x}~dx&=\frac12\int_{-\infty}^{+\infty}\frac{e^{-x^2}}{1+a\cos x}~dx\\ &=\frac12\sum_{n=-\infty}^{+\infty}\int_{2\pi n}^{2\pi+2\pi n}\frac{e^{-x^2}}{1+a\cos x}~dx\\ &=\frac12\sum_{n=-\infty}^{+\infty}\int_{0}^{2\pi}\frac{e^{-(x+2\pi n)^2}}{1+a\cos (x+2\pi n)}~dx\\ &=\frac12\sum_{n=-\infty}^{+\infty}\int_{0}^{2\pi}\frac{e^{-x^2-2\pi n x-4\pi^2n^2}}{1+a\cos x}~dx\\ &=\frac12\int_0^{2\pi}\frac{e^{-x^2}\sum_{n=-\infty}^{+\infty}e^{-2\pi n x-4\pi^2n^2}}{1+a\cos x}~dx \end{align}$$

Next I will use the result of this. After some transforms we get (note that there is a little typo in the question: in the second parameter of theta function the denominator should be equal to $\alpha$ but not $2\alpha$)

$$ I=\frac1{4\sqrt{\pi}}\int_0^{2\pi}\frac{\vartheta_3\left(\frac x2,e^{-\frac14}\right)}{1+a\cos x}~dx $$ where $\vartheta_3(u,q)$ is one of Jacobi theta functions. I'd like to note that this result impressed me a lot because of it quite simple form.

Well, let's continue. It's known that theta function could be expressed as a sum: $\vartheta_3(u,q)=1+2\sum_{n=1}^\infty q^{n^2}\cos(2 n u)$. Thus

$$\begin{align} I&=\frac1{4\sqrt{\pi}}\int_0^{2\pi}\frac{1+2\sum_{n=1}^\infty e^{-\frac14 n^2}\cos(n x)}{1+a\cos x}~dx\\ &=\frac1{4\sqrt{\pi}} \left( \int_0^{2\pi}\frac{dx}{1+a\cos x}+2\sum_{n=1}^\infty e^{-\frac14 n^2}\int_0^{2\pi}\frac{\cos(nx)}{1+a \cos(x)}~dx \right)\\ &=\frac1{4\sqrt{\pi}} \left( \frac{2\pi}{\sqrt{1-a^2}}+2\sum_{n=1}^\infty e^{-\frac14 n^2}\frac{2\pi(-1)^n \alpha ^n}{\sqrt{1-a^2}} \right)\\ &=\frac12\sqrt{\frac\pi{1-a^2}}\left(1+2\sum_{n=1}^\infty (-1)^n e^{-\frac{n^2}4}\alpha^n\right) \end{align}$$

where $\alpha=\frac1a-\sqrt{\frac1a^2-1}$.

I tried to find closed form of this infinite sum using theta functions but haven't succeeded so far.

  • $\begingroup$ This seems pretty awesome! Have you checked numerically? $\endgroup$ – Yuriy S Nov 9 '18 at 8:16
  • $\begingroup$ @Yuriy S... Yes, made it via Mathematica. All three forms (initial integral, integral with theta function, and series) agree. =)) $\endgroup$ – Mikalai Parshutsich Nov 9 '18 at 8:18
  • $\begingroup$ Is it enough to accept my answer? =^_^= $\endgroup$ – Mikalai Parshutsich Nov 9 '18 at 19:29
  • 1
    $\begingroup$ I usually wait a litte before accepting an answer, but sure $\endgroup$ – Yuriy S Nov 9 '18 at 20:14

This is not a full answer, but may be useful for other users who are interested in this integral. Basically, we are getting back to the series, linked in the other question.

Let's try the series expansion valid for $|a|<1$:

$$\int_0^\infty \frac{e^{-x^2}}{1+a \cos x}dx=\sum_{n=0}^\infty (-1)^n a^n \int_0^\infty e^{-x^2} \cos^n x~dx$$

While I don't know any general expression for the integrals inside, using ' Lord Shark the Unknown's advice, we can certainly compute the first few terms, using:

$$\int_0^\infty e^{-x^2} \cos n x~dx=\frac{\sqrt{\pi}}{2} e^{-n^2/4}$$

See also Expand $\cos^n (x)$ in terms of $\cos{kx}$, $k=1,\dots,n$. which should help to find the general terms of the series (though it would be complicated).

For the first few terms we can explicitly write:

$$\cos^0 x=1 \\ \cos^1 x= \cos x \\ \cos^2 x=\frac{1}{2} (1+\cos 2x) \\ \cos^3 x=\frac{1}{4} (3 \cos x+\cos 3x) \\ \cos^4 x=\frac{1}{8} (3+4 \cos 2x+\cos 4x) \\ \cos^5 x=\frac{1}{16} (10 \cos x+5 \cos 3x+\cos 5x)$$

So we can bound the value of the integral (for $a>0$) by the partial sums:

$$S_4=\sum_{n=0}^4 (-1)^n a^n \int_0^\infty e^{-x^2} \cos^n x~dx$$


$$S_5=\sum_{n=0}^5 (-1)^n a^n \int_0^\infty e^{-x^2} \cos^n x~dx$$

We have:

$$\int_0^\infty e^{-x^2} dx= \frac{\sqrt{\pi}}{2}\\ \int_0^\infty e^{-x^2} \cos x dx= \frac{\sqrt{\pi}}{2} e^{-1/4} \\ \frac{1}{2} \int_0^\infty e^{-x^2}(1+\cos 2x)dx=\frac{\sqrt{\pi}}{4} (1+e^{-1}) \\ \frac{1}{4} \int_0^\infty e^{-x^2}(3 \cos x+\cos 3x)dx=\frac{\sqrt{\pi}}{8} (3 e^{-1/4}+e^{-9/4}) \\ \frac{1}{8} \int_0^\infty e^{-x^2} (3+4 \cos 2x+\cos 4x) dx=\frac{\sqrt{\pi}}{16} (3+4 e^{-1}+e^{-4}) \\ \frac{1}{16} \int_0^\infty e^{-x^2} (10 \cos x+5 \cos 3x+\cos 5x) dx=\frac{\sqrt{\pi}}{32} (10 e^{-1/4}+5 e^{-9/4}+e^{-25/4})$$

We have:

$$S_5(a) < I(a) < S_4(a), \qquad 0<a<1$$

Here's the plot (scaled by $2/\sqrt{\pi}$ for convenience):

enter image description here

A much better approximation would be:

$$I(a) \approx \frac{S_4(a)+S_5(a)}{2}$$

which is a useful general trick for alternating series.

  • $\begingroup$ @AtulVerma, if you are interested in the series approximation, look up this answer ^ $\endgroup$ – Yuriy S Nov 2 '18 at 12:00

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