I tried to solve the following integral using Maple as well as by hand but unable to do so. Can anybody help me in solving the following integral?
$$ \int_{0}^{R} D\pi r^2 (D\pi r^2-1)^B 2\pi \lambda \alpha r e^{-\pi r^2(\alpha \lambda - D ln(Y))} dr $$
where $D$, $B$, $Y$, $\alpha$, $\lambda$ are constants
$\textbf{to long for comment}$ $$ \int_{0}^{R} D\pi r^2 (D\pi r^2-1)^B 2\pi \lambda \alpha r e^{-\pi r^2(\alpha \lambda - D ln(Y))} dr $$ lets tidy up the constants here $$ \lambda_1 = D\pi,\\ \lambda_2 = 2\pi\lambda\alpha,\\ \lambda_3 = \pi\alpha\lambda = \frac{\lambda_2}{2},\\ \lambda_4 = \pi(\alpha \lambda - D ln(Y)). $$ leading to an integral $$ \int\lambda_1r^2\left(\lambda_1r^2-1\right)^B\lambda_2r\mathrm{e}^{-\lambda_4r^2}dr $$
now use a substituion of the form $u = \lambda_1 r^2-1$ this yields $$ \int (u+1)u^B\lambda_2\mathrm{e}^{-\lambda_4\frac{1+u}{\lambda_1}}\frac{1}{2\lambda_1}du =\frac{\lambda_2}{2\lambda_1}\mathrm{e}^{-\frac{\lambda_4}{\lambda_1}}\int (u+1)u^B\mathrm{e}^{-\frac{\lambda_4}{\lambda_1}u}du $$ now if $B$ was integer values then that would be striaght forward by $$ I(a) = \int \mathrm{e}^{ax}dx\rightarrow I'(a) = \int x\mathrm{e}^{ax} dx \rightarrow ... $$ but alas if it is not integer then you have to just solve it :)