Question: How to solve the following integral?
$$I = \int_0^\infty \dfrac{x^{N_a + N_b – 1}}{(p \Omega_1 + \Omega_2 x)^{N_a + 1}} \ln (1 + Qx) \, _2F_1\left( N_b + 1, N_b; N_b +1; \dfrac{-\Omega_3}{\Omega_4}x\right)dx, \tag{1}$$
where $N_a, N_b \in \mathbb Z_+$ and $\Omega_1, \Omega_2, \Omega_3, \Omega_4, p, Q \in \mathbb R_+$. I am looking for a closed-form solution for the above integral. A solution in terms of any special function will also be good enough.
Any leads appreciated.
My attempt: Representing $\log(1 + Qx)$ in terms of Meijer's $G$ function, we have
\begin{align}
I = {} & \, \Omega_2^{-(N_a + 1)}\int_0^{\infty}x^{N_d + N_b – 1}\left( x + \dfrac{p\Omega_1}{\Omega_2}\right)^{-(N_a + 1)}G_{2, 2}^{1, 2}\left( Qx \left\vert \begin{smallmatrix} 1, & 1\\ 1, & 0\end{smallmatrix}\right.\right) \\
& \hspace{6cm}\times\, _2F_1\left( N_b + 1, N_b; N_b +1; \dfrac{-\Omega_3}{\Omega_4} x\right) \, dx \tag{2}
\end{align}
A solution to (2) exists in [1, eqn. 2.2], resulting into infinite series summation.
Can anyone suggest any alternate solution that doesn't contain infinite series sum?
Best Answer
Too long for a comment. The hypergeometric function inside the integral is actually elementary because of parameters: $_2F_1(\ldots)=\left(1+\frac{\Omega_3}{\Omega_4}x\right)^{-N_b}$. The integral can therefore be reduced to the form \begin{align} \int_{0}^{\infty}\left(\frac{x}{\alpha+x}\right)^{N_a+1}\left(\frac{x}{\beta+x}\right)^{N_b}\frac{\ln(1+Qx)}{x^2}\,dx=\tag{1}\\ =\int_{0}^{\infty}(1+\alpha y)^{-N_a-1}(1+\beta y)^{-N_b}\ln(1+Qy^{-1})\,dy.\tag{2} \end{align}
Now, if $N_a$, $N_b$ were arbitrary (admissible) complex numbers and if we could replace $\ln(1+Qx)$ by $(1+Qx)^s$ with some $s$, then the integral could be expressed in terms of Lauricella function $F_D^{(2)}$. When we have logarithm instead, we could write the result as a derivative of this function w.r.t. to parameter $s$ evaluated at $s=0$.
For arbitrary but fixed integer $N_a$, $N_b$ the integrals (1)-(2) look computable to me (e.g. doing partial fraction expansion first). I would not expect something more complicated than dilogarithms. Moreover for $\alpha=\beta$ the result is clearly an elementary function (integration by parts).