# Laplace transform of a product of Modified Bessel Functions

Working with a scalar field in 2 dimensions I've come to the following integral, from which I can extract the proper ultraviolet behavior ($a \ll 1$) of the theory:

$\int_0^\infty e^{-(4+a^2)x}\left[I_0(2x)\right]^2 ds$.

It is obvious to me that this is the Laplace transform of $\left[I_0(2x)\right]^2$ evaluated at $s = (4+a^2)$. From Wikipedia I got the formula

$\int_0^\infty e^{-sx} f(x)g(x) dx = \frac{1}{2\pi i} \lim_{T \to \infty} \int_{c-iT}^{c+it} F(\sigma)G(s-\sigma) d\sigma$,

where $F(\sigma)$ and $G(\sigma)$ are the Laplace transforms of $f(x)$ and $g(x)$, respectively.

I'm encountering some trouble trying to get an analytical result for that integral. What I actually need is its $a \approx 0$ behavior, but a full analytical answer would be great.

Thanks in advance for any help!

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I have used Mathematica, I must confess, but this integral is known in a closed form $$\int_0^\infty dx e^{-kx}[I_0(2x)]^2=\frac{2}{k\pi}K\left(\frac{16}{k^2}\right).$$ The condition $k>4$ must hold. Here $K(m)$ is an elliptic integral given by $$K(m) = \int_0^{\pi/2} \frac{d\theta}{\sqrt{1-m \sin^2\theta}} = \int_0^1 \frac{dt}{\sqrt{(1-t^2)(1-m t^2)}}.$$

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That's nice, thanks! But do you have any idea on how to prove it? I don't really need the proof, but it definately wouldn't hurt :D –  Felipe Attanasio Jul 27 '12 at 18:21
I think Sasha's answer fits the bill. –  Jon Jul 27 '12 at 19:07
You should edit your definition of the complete elliptic integral. Sasha's answer has the one being used by Mathematica. –  Ｊ. Ｍ. Jul 29 '12 at 5:17
@J.M.: Thanks for your comment. I did it taking into account that I worked this out with Mathematica and held the further definition taken Wikipedia. –  Jon Jul 29 '12 at 9:55

Given that $I_0(2 x)^2 = 1 + 2 x^2 + \frac{3}{2} x^4 + \frac{5}{9}x^6 + \mathcal{o}(x^6)$ we see that it is a hypergeometric function: $$I_0(2x)^2 = {}_1F_2\left(\frac{1}{2}; 1,1; 4 x^2\right) = \sum_{n=0}^\infty \frac{\left(\frac{1}{2}\right)_n}{(1)_n (1)_n} \frac{(4 x^2)^n}{n!} = \sum_{n=0}^\infty \left(\frac{x^{n}}{n!} \right)^2 \binom{2n}{n}$$ Now, integrate term-wise: $$\int_0^\infty \mathrm{e}^{-k x} [ I_0(2x) ]^2 \mathrm{d} x = \sum_{n=0}^\infty \frac{(2n)!}{n!^4} \int_0^\infty x^{2n} \mathrm{e}^{-k x} \mathrm{d} x = \sum_{n=0}^\infty \frac{(2n)!}{n!^4} \frac{(2n)!}{k^{2n+1}} = \frac{1}{k} \sum_{n=0}^\infty \left( \binom{2n}{n} \frac{1}{k^n} \right)^2$$ The sum is again hypergeometric, since ratio of subsequent summands is $\frac{4 (2n+1)^2}{k^2 (n+1)^2} = \frac{16}{k^2} \frac{\left( n+ \frac{1}{2}\right)^2}{(n+1)^2}$, thus the sum equals: $$\int_0^\infty \mathrm{e}^{-k x} [ I_0(2x) ]^2 \mathrm{d} x = \frac{1}{k} \cdot {}_2F_1\left( \frac{1}{2}, \frac{1}{2}; 1; \frac{16}{k^2}\right) = \frac{2}{ \pi k} K\left(\frac{16}{k^2}\right)$$ where $K(m)$ is a complete elliptic integral: $$K(m) = \int_0^{\pi/2} \frac{\mathrm{d} \phi}{\sqrt{1-m \cdot \sin^2(\phi)}}$$

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Whoa! Thanks a lot! –  Felipe Attanasio Jul 27 '12 at 19:47
@FelipeAttanasio: so you should consider accepting this answer. –  enzotib Jul 27 '12 at 21:38