I am trying to solve an equation for my paper and I am stuck at the following point beyond which I cannot go. I will appreciate any help regarding finding the closed form expression.
$$R = \int_{x>0} \sum_{j=0}^{40}\sum_{m=0}^{j}\frac{K^j(1-e^x)^{j-m}}{e^K.j!(j-m)!}\sum_{q=0}^{40}\frac{(-1)^qA^q}{q!}(e^x-1)^{2q/\alpha - (j-m)} \frac{\Gamma(2q/\alpha+1)}{\Gamma(2q/\alpha-(j-m) + 1)} dx$$
Note that $A$ is further a variable which depends on several variables and in this particular equation where we integrate w.r.t $x$, $A$ is treated as a constant. $\Gamma(.)$ is the Gamma function
Edit: First of all to simplify let me assume:
$A_{\Gamma} = \frac{\Gamma(2q/\alpha+1)}{\Gamma(2q/\alpha-(j-m) + 1)}$
$$R = \int_{x>0} \sum_{j=0}^{\infty}\sum_{m=0}^{j}\frac{K^j(-1)^{j-m}(e^x-1)^{j-m}}{e^K.j!(j-m)!}\sum_{q=1}^{\infty}\frac{(-1)^qA^q}{q!}(e^x-1)^{2q/\alpha - (j-m)} A_\Gamma\ dx$$
$$R = \int_{x>0} \sum_{j=0}^{\infty}\sum_{m=0}^{j}\sum_{q=1}^{\infty}\frac{K^j(-1)^{j-m+q}(e^x-1)^{j-m + 2q/\alpha - (j -m)}}{e^K.j!(j-m)!}\frac{A^q}{q!} A_\Gamma\ dx$$
$$R = \int_{x>0} \sum_{j=0}^{\infty}\sum_{m=0}^{j}\sum_{q=1}^{\infty}\frac{K^j(-1)^{j-m+q}(e^x-1)^{2q/\alpha}}{e^K.j!(j-m)!}\frac{A^q}{q!} A_\Gamma\ dx$$
Hence, to simplify, first I have to fine $f(x) = \int_0^\infty(e^x-1)^{2q/\alpha} dx$
$$R = \int_{x>0} \exp{(e^x-1)^{2/\alpha}}\sum_{j=0}^{\infty}\sum_{m=0}^{j}\sum_{q=1}^{\infty}\frac{K^j(-1)^{j-m+q}}{e^K.j!(j-m)!}\frac{A^q}{q!} A_\Gamma\ dx$$