I'm trying to solve the following integral $$ T=\int_0^\infty \exp(-a x^2) I_1(b x) \log(I_1(b x))\, dx $$ where $I_1(x)$ is the modified Bessel function of the first kind and order one, and $a$, $b$ are positive real constants.

Any ideas on how to tackle this problem?


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    $\begingroup$ If $I(k)=\displaystyle\int_0^\infty e^{-ax^2}I_1^k(bx)~dx~$ then $T=I'(1)$. $\endgroup$ – Lucian Feb 4 '15 at 11:52

We can get rid of a parameter by setting $x=\frac{1}{\sqrt{a}}z$, then, provided that $\lambda=\frac{b}{\sqrt{a}}$:

$$ I(\lambda) = \frac{1}{\sqrt{a}}\int_{0}^{+\infty}\exp\left(-z^2\right) I_1(\lambda z)\log\left(I_1(\lambda z)\right)\, dz. $$ Now $I_1(x)$ is a solution of the differential equation: $$ x^2 f''+ xf'=(x^2+1)\,f, $$ hence it follows that: $$ \lambda^2 I''(\lambda)+\lambda\, I'(\lambda) = \frac{1}{\sqrt{a}}\int_{0}^{+\infty}e^{-z^2}\left(\lambda^2 z^2\frac{I_1'(\lambda z)^2}{I_1(\lambda z)}+(1+\log I_1(\lambda z))(1+\lambda^2 z^2)I_1(\lambda z)\right)\,dz $$ Now the plan is to get rid of everything computable in the right hand side of the last expression, in order to have $I(\lambda)$ as a solution of a non-homogeneous ODE. This looks pretty hard, however. Partial results are: $$ \int_{0}^{+\infty}e^{-z^2}I_1(\lambda z)\,dz = \frac{1}{\lambda}\left(e^{\frac{\lambda^2}{4}}-1\right),$$ $$ \int_{0}^{+\infty}z^2 e^{-z^2}I_1(\lambda z)\,dz = \frac{\lambda}{4}e^{\frac{\lambda^2}{4}},$$ they are easily achieved through the (inverse) Laplace transform.

By the way, the original integral depends on the Kullback-Leibler divergence between a strange distribution, having $\frac{1}{e^{1/4}-1}e^{-x^2}I_1(x)$ as a pdf supported on $\mathbb{R}^+$, and a normal distribution.


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