I have a very direct question.
Is there an explicit solution for this integral:
$$\int_a^b e^{-\frac{x^2}{2\sigma^2}}I_0\left(\frac{x\cdot z}{\sigma^2}\right)\mathrm{d}x$$
where $z,\sigma^2\in\mathbb{R}^+$ and $I_0$ is the 0-order modified Bessel function of first kind?
Thanks in advance.
Best Answer
$$J=\int e^{-\frac{x^2}{2\sigma^2}}\,I_0\left(\frac{x\, z}{\sigma^2}\right)\,dx$$
Let $x=\frac{\sigma ^2 y}{z}$ and $k=\frac{\sigma ^2}{2 z^2}$ to make $$J=\frac {\sigma ^2 }{z} \int e^{-k y^2}\,I_0(y)\,dy$$
Now using the infinite series $$I_0(y)=\sum_{n=0}^\infty \frac{2^{-2 n}}{(n!)^2}\, y^{2n}$$
we face a series of integrals $$J_n= \int e^{-k y^2}\,y^{2n}\,dy=-\frac{1}{2} k^{-\frac{2n+1}{2}} \Gamma \left(n+\frac{1}{2},k y^2\right)$$
For example, let $z=\frac 12$ and $\sigma=\frac 1{\sqrt 2}$ to make the simple $$J=\int_0^1 e^{-x^2}\, I_0(x)\,dx=\sum_{n=0}^\infty \frac{ \Gamma \left(n+\frac{1}{2},0\right)-\Gamma \left(n+\frac{1}{2},1\right)}{2^{2 n+1}\,(n!)^2}$$ which converge very fast. For the partial sums up to $n=p$ $$\left( \begin{array}{cc} p & \text{approximation} \\ 0 & 0.7468241328 \\ 1 & 0.7941922193 \\ 2 & 0.7957589192 \\ 3 & 0.7957878829 \\ 4 & 0.7957882194 \\ 5 & 0.7957882221 \\ 6 & 0.7957882221 \end{array} \right)$$