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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!

2 Answers 2

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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!2012-07-27
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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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    @J.M.: Than$k$s for your comment. I did it ta$k$ing into account that I wor$k$ed this out with Mathematica and held the further definition ta$k$en Wikipedia.2012-07-29