# Analytical closed form for a definite integral

I am doing a computation in quantum field theory and the following integral occurred to me $$I(a)=\int_{-\infty}^{+\infty}\frac{e^{-a\sqrt{x^2+1}}dx}{x^2+1} \qquad a\ge 0.$$ I would like to know a closed form for it and the eventual steps to achieve the result.

Thanks.

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## 3 Answers

Let me denote your integral by $I(a)$. Making the change of variables $x=\sinh\varphi$, we get $$I(a)=\int_{-\infty}^{\infty}\frac{e^{-a\cosh\varphi}d\varphi}{\cosh\varphi}.$$ Next, differentiating this expression with respect to $a$, we get $$I'(a)=-\int_{-\infty}^{\infty}e^{-a\cosh\varphi}d\varphi=- 2K_0(a),$$ where $K_0(a)$ denotes the Macdonald function. Integrating back, we find that $$I(a)=I(0)-2\int_0^aK_0(x)dx=\pi-2\int_0^aK_0(x)dx.$$ According to Prudnikov-Brychkov-Marychev (Vol.2, formula 1.12.1.4), the remaining integral is expressed in terms of Macdonald and modified Struve functions, so that $$I(a)=\pi-2a\left\{K_0(a)+\frac{\pi}{2}\left[K_0(a)\mathbf{L}_1(a)+K_1(a)\mathbf{L}_0(a)\right]\right\}.$$ Presumably this answer does not add anything useful to your integral except that the corresponding special functions have a name, but at least one can be sure that it cannot be simplified further.

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Thanks a lot. This is simply perfect! These functions are very well-known in their behaviors and it is what I need. – Jon Sep 11 '13 at 14:15

Here is a closed form in terms of the Meijer $G$-function

$$I(a)=\int_{-\infty}^{+\infty}\frac{e^{-a\sqrt{x^2+1}}dx}{x^2+1} =\frac{a^2}{4}\, G^{3, 0}_{1, 3}\left(\frac{{a}^{2}}{4}\, \Bigg\vert\,^{0}_{-\frac{1}{2}, -\frac{1}{2}, -1}\right).$$

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Mathematica seems to disagree with this result. It is possible that my interpretation is not correct. This is my code (a^2/4)*MeijerG[{{}, {0}}, {{-1/2, -1/2, -1}, {-1/2, -1/2, -1}}, a^2/4]. – Jon Sep 11 '13 at 14:05
@Jon: It should be the same answer. I believe Mathematica uses the second definition in this link while I used the first as you can see from the link in my answer. – Mhenni Benghorbal Sep 11 '13 at 21:23

There is a mistake in the original post: thanks to Jon for pointing it out.

Recognize that the function is even, so you may integrate from 0 to $+\infty$ and multiply by 2, substitute $u=\sqrt{x^2+1}$, and you are left with $$\int_1^{\infty} \frac{e^{-au}}{u^2}\sqrt{u^2-1}du$$

Which still will not have a closed form involving elementary functions.

Original post:

Recognize that the function is even, so you may integrate from 0 to $+\infty$ and multiply by 2, substitute $u=\sqrt{x^2+1}$, and you are left with $$\int_1^{\infty} \frac{e^{-au}}{u^2}=2E_2(a).$$

Where $E_2(\cdot)$ is the generalized exponential integral, so a closed form does not exist.

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It does not seem so straightforward to substitute $u=\sqrt{x^2+1}$ as this gives $udu=xdx$. You cannot get rid of that $x$. – Jon Sep 11 '13 at 13:04
yes, there is a mistake. – Mhenni Benghorbal Sep 11 '13 at 13:05