Integration of hypergeometric function involving power and exponential function

Question

I would be grateful if someone can explain how to evaluate the following $$ \int_{0}^{\infty} y^a exp(-y/b)\,_2F_1(1,1;2;-c y^2) dy $$ where $a>0$, $b>0$, and $c>2$. I searched on special functions (using Gradestien book and others) but I could not find an answer.

Answer 1

I do not know if a closed form exist.

Let $y=\frac{x}{\sqrt{c}}$ and $k=\frac{1}{b \sqrt{c}}$ to make the integral to be $$ c^{-\frac{a+1}{2} }\int_{0}^{\infty}x^{a-2} e^{-k x} \log \left(x^2+1\right)\,dx$$ Let $$J_a=\int_{0}^{\infty}x^{a-2} e^{-k x} \log \left(x^2+1\right)\,dx$$ For $J_2$ it is simple. Now appear the Meijer G function in all the cases I tried $$J_1=\frac{G_{2,4}^{4,1}\left(\frac{k^2}{4}| \begin{array}{c} 0,1 \\ 0,0,0,\frac{1}{2} \end{array} \right)}{2 \sqrt{\pi }}\qquad \qquad J_3=\frac{2 G_{1,3}^{3,1}\left(\frac{k^2}{4}| \begin{array}{c} 0 \\ 0,0,\frac{3}{2} \end{array} \right)}{\sqrt{\pi } k^2}$$

$$J_4=\frac{4 G_{2,4}^{4,1}\left(\frac{k^2}{4}| \begin{array}{c} 0,1 \\ 0,0,\frac{3}{2},2 \end{array} \right)}{\sqrt{\pi } k^3}\qquad \qquad J_5=\frac{8 G_{2,4}^{4,1}\left(\frac{k^2}{4}| \begin{array}{c} 0,1 \\ 0,0,2,\frac{5}{2} \end{array} \right)}{\sqrt{\pi } k^4}$$

$$J_6=\frac{16 G_{2,4}^{4,1}\left(\frac{k^2}{4}| \begin{array}{c} 0,1 \\ 0,0,\frac{5}{2},3 \end{array} \right)}{\sqrt{\pi } k^5}\qquad \qquad J_7=\frac{32 G_{2,4}^{4,1}\left(\frac{k^2}{4}| \begin{array}{c} 0,1 \\ 0,0,3,\frac{7}{2} \end{array} \right)}{\sqrt{\pi } k^6}$$

$$J_8=\frac{64 G_{2,4}^{4,1}\left(\frac{k^2}{4}| \begin{array}{c} 0,1 \\ 0,0,\frac{7}{2},4 \end{array} \right)}{\sqrt{\pi } k^7}\qquad \qquad J_9=\frac{128 G_{2,4}^{4,1}\left(\frac{k^2}{4}| \begin{array}{c} 0,1 \\ 0,0,4,\frac{9}{2} \end{array} \right)}{\sqrt{\pi } k^8}$$ and, I hope, we could conjecture that for $a >3$ $$\color{blue}{J_a=\frac {2^{a-2}}{k^{a-1}\,\sqrt \pi}G_{2,4}^{4,1}\left(\frac{k^2}{4}| \begin{array}{c} 0,1 \\ 0,0,\frac{k-1}2,\frac{k}{2} \end{array} \right)}$$