# Integral involving hypergeometric functions - $\int_{0}^{\infty} e^{(ax)} U(c,d,b_1x)U(c,d,b_2x) dx$

I want to evaluate the improper integral of $\int_{0}^{\infty} e^{(ax)} U(c,d,b_1x)U(c,d,b_2x) \, dx$ where U is the confluent hypergeometric function, $d$ is a real number and $d \in (1,2)$, $a$ and $c$ are complex numbers with both real parts $\Re(a) >0$ and $\Re(b_1), \Re(b_2) < 0$

Here is my attempt:

1. For starters, I tried to evaluate $\int_{0}^{\infty} e^{(x)} U(1,1,x) \, dx$ (a simple case) in Wolfram, there seems to be closed form solution. The indefinite integral and plots for various values of $(a,b)$ are here. However, I am unable to compute $\int_{0}^{\infty} e^{(x)} U(1,1,x) U(1,1,x) \, dx$ even with Wolfram.

If I can understand the step-by-step process of evaluating the integral for the simple case, I should be able to evaluate even for the general case, but unfortunately, Wolfram does not provide it.

I would be thankful if someone can help me with this integration of $$\int_{0}^{\infty} e^{(ax)} U(c,d,b_1x)U(c,d,b_2x) \, dx$$

• mathoverflow.net/a/210897/118924 - A similar question. From this, I think we can evaluate integrals of the type $\int_{0}^{\infty} e^{(ax)} U(c,d,b_1x) \, dx$ – kasa Apr 14 '18 at 18:27
• Are the real parts of $b_{1}$ and $b_{2}$ both greater than zero as well? If so, I think your improper integral diverges for $0\le\Re{\left(a+b_{1}+b_{2}\right)}$. – David H Apr 15 '18 at 9:03
• @DavidH Thank you for pointing it out. Yes, $0\ge\Re (b_1), \Re(b_2)$. I am editing the original question too. – kasa Apr 15 '18 at 14:05
• @DavidH We can consider only the case of $d \in (1,2)$, if it may help. – kasa Apr 15 '18 at 14:38
• @DavidH Sure, I too will look into that approach. Does it help to consider that $b_{1,2}$ are just -ve real numbers, instead of considering them to be complex? – kasa Apr 15 '18 at 15:26