Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

3 Answers 3

up vote 2 down vote accepted

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.

share|improve this answer
    
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

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

share|improve this answer
    
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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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