2
$\begingroup$

My goal is to integrate the following function: $$ \int_0^{\infty } x^a \exp \left(-\frac{c x^2+f x}{b}\right) \, dx $$ where, $a, b, c > 0$ and $a, b, c, f \in \mathbb{R}$.

Mathematica gives me the answer: $$ \frac{\left(\frac{c}{b}\right)^{-\frac{a}{2}} \left(-f \Gamma \left(1+\frac{a}{2}\right) \, _1F_1\left(1+\frac{a}{2};\frac{3}{2};\frac{f^2}{4 b c}\right)+b \sqrt{\frac{c}{b}} \Gamma \left(\frac{1+a}{2}\right) \, _1F_1\left(\frac{1+a}{2};\frac{1}{2};\frac{f^2}{4 b c}\right)\right)}{2 c} $$ but I don't understand how it obtains this result, or where Kummer's confluent hypergeometric $_1F_1$ comes from. I know some of the integral representations of $_1F_1$, but those range usually integrate from 0 to 1. Can anybody explain how Mathematica solves this? Thanks in advance!

$\endgroup$
2
  • $\begingroup$ Using $d$ as a parameter in integrals is not very convenient, especially when you have $dx$ and $dx$ have different meanings :) As for the function, you could try substitution in the integral to get it into the desired form $\endgroup$
    – Yuriy S
    Oct 26, 2018 at 10:40
  • $\begingroup$ Sorry about that, I changed it to $f$. What substitution do you mean? $\endgroup$
    – Vandenman
    Oct 26, 2018 at 11:40

1 Answer 1

2
$\begingroup$

Let $I(a)$ denote your integral. Integration by parts gives $$ \begin{align*} I(a) & = -\int_{0}^{\infty} \frac{x^{a+1}}{a+1}(-\frac{2cx + f}{b})\exp(-\frac{cx^2 + fx}{b}) \\ & = \frac{2c}{b(a+1)}I(a+2) + \frac{f}{b(a+1)}I(a+1) \end{align*} $$ or, in other terms, $$ I(a+2) = - \frac{f}{2c} I(a+1) + \frac{b(a+1)}{2c}I(a) $$

I assume that Mathematica tries to solve this recurrence (computing $I(0)$ and $I(1)$ along the way) and ends up with hypergeometric functions because they satisfy many.

$\endgroup$
2
  • 1
    $\begingroup$ It's a very good way to find the integral, but Mathematica doesn't know that. I'd guess it uses Meijer G functions and reduction formulas for that kind of integrals $\endgroup$
    – Yuriy S
    Oct 30, 2018 at 9:56
  • $\begingroup$ I was unaware of the existence of such functions, thank you for pointing that out! $\endgroup$
    – Ϛ .
    Oct 30, 2018 at 13:14

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .