I am trying to find the closed form expression of the following equation for my paper, but not getting it correctly:
$$P_{o} = \int_0^{\infty}\frac{1}{\sqrt{\alpha^2\sigma_f^2\sigma_h^2}}\times\sum_{m=0}^{\infty}{\sigma_r}W_{-m-0.5,0}(2z)\left[e^z\left[\Gamma(1+m)-\Gamma\left(1+m,\frac{\left(\gamma_s(1+y)-1\right)}{A\sigma^2_r}\right)\right]\right]\frac{1}{B}\times\sum_{k=0}^{\infty}\frac{\left(\frac{y}{B}\right)^k}{\sigma^{2k+1}_e}W_{-k-0.5,0}(2\zeta)\frac{1}{\sqrt{\alpha^2\sigma_f^2\sigma_g^2}}e^{\zeta-\frac{y}{B\sigma^2_e}}\text{d}y$$
where $W_{\lambda,\mu}(\cdot)$ is Whittaker function, $y$ is integration variable and all remaining are constants.
I will appreciate any help regarding finding the closed form expression.
Try to factor out constants like the ones not in terms of the sum’s index. $$\sum_{k=0}^{\infty}\frac{\left(\frac{y}{B}\right)^k}{\sigma^{2k+1}_e}W_{-m-0.5,0}(2\zeta)\frac{1}{\sqrt{\alpha^2\sigma_f^2\sigma_g^2}}e^{\zeta-\frac{y}{B\sigma^2_e}}=\frac{1}{\sqrt{\alpha^2\sigma_f^2\sigma_g^2}}e^{\zeta-\frac{y}{B\sigma^2_e}} \sum_{k=0}^{\infty} W_{-k-0.5,0}(2\zeta)\frac{\left(\frac{y}{B}\right)^k}{\sigma^{2k+1}_e}= \frac{1}{\sigma_e\sqrt{\alpha^2\sigma_f^2\sigma_g^2}}e^{\zeta-\frac{y}{B\sigma^2_e}} \sum_{k=0}^{\infty} W_{-k-\frac12,0}(2\zeta)\,\left(\frac{y}{B\sigma^{2}_e}\right)^k $$
The Whittaker function cannot be simplified, only in terms of other hypergeometric functions for now. However, I did find this integral representation for the Whittaker function. Note the constant terms in the integrand:
$$\text W_{v,u}(z)\mathop=^{\text{Re}(z)>0}_{\text{Re}(u-v)>-\frac12}\int_{\frac12}^\infty \frac{z^{u+\frac12}}{Γ\left(u-v+\frac12\right)}e^{-tz} \left(t-\frac12\right)^{u-v-\frac12}\left(t+\frac12\right)^{u+v-\frac12}dt$$
Therefore: $$\text W_{-k-\frac12,0}(2ζ)\mathop=^{\text{Re}(ζ)>0}_{\text{Re}\big(k+\frac12\big)>-\frac12} \int_{\frac12}^\infty \frac{\sqrt 2ζ^{\frac12}}{k!}e^{-2tζ} \left(t-\frac12\right)^{k}\left(t+\frac12\right)^{-k-1}dt $$
I will ignore everything except the $k$ indexed sum. Assuming that ${\text{Re}(ζ)>0},{\text{Re}\big(k+\frac12\big)>-\frac12}$, we have the following, but $k\in\Bbb N^0$, so we can just assume $\text {Re}(ζ)>0$. Let’s switch operators integrating term by term and factor:
$$\sum_{k=0}^{\infty} W_{-k-\frac12,0}(2\zeta)\,\left(\frac{y}{B\sigma^{2}_e}\right)^k = \sum_{k=0}^{\infty} \int_{\frac12}^\infty \frac{\sqrt 2ζ^{\frac12}}{k!}e^{-2tζ} \left(t-\frac12\right)^{k}\left(t+\frac12\right)^{-k-1}\,\left(\frac{y}{B\sigma^{2}_e}\right)^k dt= \sqrt 2ζ^{\frac12}\int_{\frac12}^\infty e^{-2tζ}\sum_{k=0}^{\infty} \frac{1}{k!}\left(t-\frac12\right)^{k}\left(t+\frac12\right)^{-k-1}\,\left(\frac{y}{B\sigma^{2}_e}\right)^k dt$$
I will use machine aid for accuracy and sum evaluation:
$$\sqrt 2ζ^{\frac12}\int_{\frac12}^\infty e^{-2tζ}\sum_{k=0}^{\infty} \frac{1}{k!}\left(t-\frac12\right)^{k}\left(t+\frac12\right)^{-k-1}\,\left(\frac{y}{B\sigma^{2}_e}\right)^k dt = \sqrt 2ζ^{\frac12}\int_{\frac12}^\infty \frac{e^{\frac{(2t-1)y}{(2t+1)B\sigma^{2}_e}-2tζ}}{t+\frac12} dt $$
which has no closed form, for now. Here is your final result with $\text{Re}(ζ)>0$:
$$\sum_{k=0}^{\infty}\frac{\left(\frac{y}{B}\right)^k}{\sigma^{2k+1}_e}W_{-m-0.5,0}(2\zeta)\frac{1}{\sqrt{\alpha^2\sigma_f^2\sigma_g^2}}e^{\zeta-\frac{y}{B\sigma^2_e}}= \frac{\sqrt{2ζ} e^{\zeta-\frac{y}{B\sigma^2_e}}}{\sigma_e\sqrt{\alpha^2\sigma_f^2\sigma_g^2}} \int_{\frac12}^\infty \frac{e^{\frac{(2t-1)y}{(2t+1)B\sigma^{2}_e}-2tζ}}{t+\frac12} dt $$
The following is true using the Lower Incomplete Gamma function $γ(a,z)=Γ(a)-Γ(a,z)$. I also combined the square root parts and much factoring. Do not cancel the radical and powers as the variables with absolute value signs as they may be complex:
$$P_{o} =\frac{\sigma_r\sqrt{2ζ}e^{\zeta-z}}{B\sigma_e\sqrt{\alpha^4\sigma_f^4\sigma_g^2\sigma_h^2}}\int_0^{\infty}\sum_{m=0}^{\infty}γ\left(1+m,\frac{\gamma_s(1+y)-1}{A\sigma^2_r}\right) W_{-m-\frac12,0}(2z) e^{-\frac{y}{B\sigma^2_e}} \int_{\frac12}^\infty \frac{e^{\frac{(2t-1)y}{(2t+1)B\sigma^{2}_e}-2tζ}}{t+\frac12} dt dy$$
Try doing the same technique with integral representations of the lower gamma function and Whittaker function. You should ask how to simplify this.
Good luck finding a closed form. Please correct me and give me feedback!