How to calculate $$\int_0^\infty \frac{x}{\lambda}\cdot \exp\left(-\frac{x+\lambda}{2}\right)\cdot\frac{I_1^2(\sqrt{x\lambda})}{I_0(\sqrt{x\lambda})}\,dx,$$
where $\lambda$ is a positive constant, $I_p(\cdot)$ is the modified Bessel function of the first kind.
Some properties may be useful：
- The p.d.f. of noncentral chi square random variable with degree of freedom $n$ and noncentral parameter $\lambda$ is
$$f(x;n,\lambda)=\frac{1}{2}\left(\frac{x}{\lambda}\right)^{\frac{n-2}{4}}\cdot \exp\left(-\frac{x+\lambda}{2}\right)\cdot I_{\frac{n}{2}-1}\left(\sqrt{x\lambda}\right).$$
Thus $$\int_0^\infty \frac{x}{\lambda}\cdot \exp\left(-\frac{x+\lambda}{2}\right)\cdot\frac{I_1^2(\sqrt{x\lambda})}{I_0(\sqrt{x\lambda})}\,dx=2\int_0^\infty \frac{f^2(x;4,\lambda)}{f(x;2,\lambda)}\,dx.$$
$$\frac{dI_0(z)}{dz}=I_1(z).$$
$$\frac{dI_1(z)}{dz}=I_0(z)-\frac{I_1(z)}{z}.$$
Maybe this integral cannot be analytically solved. After consulting a lot of materials, I found this document
An integral in this paper is defined as: $$ \Psi(\rho)=\int_0^\infty\frac{y^3}{\rho^2}\exp\left(-\frac{y^2}{2\rho}-\frac{\rho}{2}\right)\frac{I_1^2(\rho)}{I_0(\rho)}dy-\rho. $$
And this paper also give the upper and lower bound, and even approximate solution of this intergal.
The upper bound is $$\Psi_u(\rho)=\min\{\rho,1\}.$$ The lower bound is $$\Psi_l(\rho)=\frac{\rho}{\rho+1}.$$ And the approximate solution is $$\Psi_a(\rho)=\frac{\pi\rho\left[I_0\left(\frac{\rho}{4}\right)+I_1\left(\frac{\rho}{4}\right)\right]^2}{8(2+\rho)e^{\frac{\rho}{2}}-\pi\left[(2+\rho)I_0\left(\frac{\rho}{4}\right)+\rho I_1\left(\frac{\rho}{4}\right)\right]^2}.$$
Thus the integral I wanted to solve can be reduced to $$ \begin{align} \int_0^\infty\frac{x}{\lambda}\exp\left(-\frac{x+\lambda}{2}\right)\frac{I_1^2\left(\sqrt{x\lambda}\right)}{I_0\left(\sqrt{x\lambda}\right)}dx &=2\int_0^\infty\frac{y^3}{\lambda^3}\exp\left(-\frac{y^2}{2\lambda}-\frac{\lambda}{2}\right)\frac{I_1^2(y)}{I_0(y)}dy\\ &=2\left[\frac{\Psi(\lambda)}{\lambda}+1\right]\\ &\approx 2\left[\frac{\Psi_a(\lambda)}{\lambda}+1\right]. \end{align} $$