0
$\begingroup$

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?

Thanks!!

$\endgroup$
  • 1
    $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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