Recently, I have encountered the following integral solution problem in my research. Because it involves special functions, I cannot successfully solve it in calculation.
$$\mathbb{E}_{Z_{1},Z_{2},\hat{h}}\log(I_{n_{\mathrm{d}}-1}\left(2 \left|\hat{h}\right| ||y|| \sqrt{n_{\mathrm{d}} \rho_{\mathrm{d}}}\right)), \quad (1)$$
where $I_{n_{d}−1}(\cdot)$ the modified Bessel function of the first kind with $n_{d}−1$ degrees of freedom, $Z_{1}$ obeys the $\Gamma(1,1)$ distribution and $Z_{2}$ obeys the $\Gamma(n_{d}-1,1)$ distribution, $||y||^{2}$ equals to $(1+n_{d}\rho_{d})Z_{1}+Z_{2}$ in the distribution sense, $n_{d}$ , $\rho_{d}$ are all known, $h$ obeys the complex Gaussian distribution $\mathbf{CN}(0,1)$ and $\hat{h}|h$ obeys the complex Gaussian distribution $\mathbf{CN}(h,1/(n_{p}\rho_{p}))$ , $|\hat{h}|$ is the norm of $h$.
At present, my preliminary idea is to deflate the above expectations according to Jensen inequality due to the convexity of the $-\log(\cdot)$ fucntion and put the expectations into logarithm.
I could get the following upper bound of the above integral:
$$\mathbb{E}_{Z_{1},Z_{2},\hat{h}}\log(I_{n_{\mathrm{d}}-1}\left(2 \left|\hat{h}\right| ||y|| \sqrt{n_{\mathrm{d}} \rho_{\mathrm{d}}}\right))\leq \log (\mathbb{E}_{Z_{1},Z_{2},\hat{h}}I_{n_{\mathrm{d}}-1}\left(2 \left|\hat{h}\right| ||y|| \sqrt{n_{\mathrm{d}} \rho_{\mathrm{d}}}\right)) ,$$
where $Z_{1},Z_{2}$ and $\hat{h}$ are independent.
Nevertheless, there is no idea about how to deal with the above integral of Bessel function.
Or if the integral $(1)$ has the closed form ？
I would be grateful if someone could help solve this problem .
Thanks, Liu .
Maybe I could calculate the above integral in the following way ?
\begin{align} &\mathbb{E}_{Z_{1},Z_{2},\hat{h}}I_{n_{d}-1}(2\hat{h}||y||\sqrt{n_{d}\rho_{d}})\\ =&\mathbb{E}_{\hat{h}}\int_{0}^{\infty}\int_{0}^{\infty}e^{-z_{1}}\frac{z_{2}^{n_{d}-2}e^{-z_{2}}}{(n_{d}-2)!}I_{n_{d}-1}(2|\hat{h}|\sqrt{n_{d}\rho_{d}}((1+n_{d}\rho_{d})z_{1}+z_{2}))dz_{1}dz_{2}, \end{align} where $z_{1}\sim \Gamma(1,1)$ and $z_{2}\sim \Gamma(n_{d}-1,1)$.
Then I calculated the double integral in the above integral.
Using the series expansion form of Bessel function, I get that \begin{align} &\int_{0}^{\infty}\int_{0}^{\infty}e^{-z_{1}}\frac{z_{2}^{n_{d}-2}e^{-z_{2}}}{(n_{d}-2)!}I_{n_{d}-1}(2|\hat{h}|\sqrt{n_{d}\rho_{d}}((1+n_{d}\rho_{d})z_{1}+z_{2}))dz_{1}dz_{2}\\ =&\int_{0}^{\infty}\int_{0}^{\infty}e^{-z_{1}}\frac{z_{2}^{n_{d}-2}e^{-z_{2}}}{(n_{d}-2)!}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n!\Gamma(n+n_{d}-1+1)}\frac{(2|\hat{h}|\sqrt{n_{d}\rho_{d}})^{2n+n_{d}-1}}{2^{2n+n_{d}-1}}\\ &((1+n_{d}\rho_{d})z_{1}+z_{2})^{n+\frac{n_{d}-1}{2}} dz_{1}dz_{2}\\ \stackrel{\mathrm{Fubini}}{=}& \sum_{n=0}^{\infty}\frac{(-1)^{n}(|\hat{h}|\sqrt{n_{d}\rho_{d}})^{2n+n_{d}-1}}{n!\Gamma(n+n_{d})(n_{d}-2)!} \int_{0}^{\infty}\int_{0}^{\infty}e^{-z_{1}}z_{2}^{n_{d}-2}e^{-z_{2}}((1+n_{d}\rho_{d})z_{1}+z_{2})^{n+\frac{n_{d}-1}{2}} dz_{1}dz_{2}. \quad (1) \end{align} Let $x=z_{1},y=z_{2},\alpha=1+n_{d}\rho_{d},\beta=n+\frac{n_{d}-1}{2},\nu=n_{d}-1$, then we may simplified the integral in the above series to the following form. \begin{align} &\int_{0}^{\infty}\int_{0}^{\infty}e^{-z_{1}}z_{2}^{n_{d}-2}e^{-z_{2}}((1+n_{d}\rho_{d})z_{1}+z_{2})^{n+\frac{n_{d}-1}{2}} dz_{1}dz_{2}\\ =& \int_{0}^{\infty}\int_{0}^{\infty}e^{-x}y^{\nu-1}e^{-y}(\alpha x+y)^{\beta}dydx\\ =& \int_{0}^{\infty}y^{\nu-1}e^{-y}\int_{0}^{\infty}e^{-x}(\alpha x+y)^{\beta}dxdy.\quad (2) \end{align} According to Table of Integrals, Series, and Products (Seventh Edition)[M]. 2007. 3.462.12， we get \begin{align} & \int_{0}^{\infty}y^{\nu-1}e^{-y}\int_{0}^{\infty}e^{-x}(\alpha x+y)^{\beta}dxdy\\ =& \int_{0}^{\infty}y^{\nu-1}e^{-y}(\alpha^{\beta}e^{\frac{y}{\alpha}}\Gamma(\beta+1,\frac{y}{\alpha}))dy,\quad (3) \end{align} where $\Gamma(\cdot,\cdot)$ represents the upper incomplete gamma function, i.e., $\Gamma(s,x)=\int_{x}^{\infty}t^{s-1}e^{-t}dt$.
Then what should I do with integral (3)?
I would be grateful if anyone could give me some advice.
Thanks a lot ! Liu.