Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This integral came up in a research problem I'm working on, but I haven't had much luck calculating it. I suspect that the integral doesn't have a very clean form, but if anyone knows of an easy substitution or some elementary way to evaluate this integral, it would be very helpful.


share|cite|improve this question
Mathematica does not find a solution at all, not even with Appell Hypergeometric functions in conjunction with error functions, I dont think you will either... I am very sorry; I am not saying its impossible, but very, VERY hard. – CBenni Jan 30 '13 at 0:03
Actually, Mathematica 9 gives an answer in terms of Bessel and generalized hypergeometric functions. Try Assuming[c>0,Integrate[x^2/(x+c)^(3/2) Exp[-x^2],{x,0,\[Infinity]}]]. – Ayman Hourieh Jan 30 '13 at 0:14
@AymanHourieh I overlooked the bounds >_< sorry. I will post the solution as an answer, because it is impossible in the comments. – CBenni Jan 30 '13 at 0:24
Of course you must assume $c \notin (-\infty, 0]$ for this to exist. – Robert Israel Jan 30 '13 at 0:25
@RobertIsrael Yes, you're absolutely right. Should have mentioned that c is a positive constant. – Sam Jan 30 '13 at 4:44

Denote $b[x]:=\; \text{BesselI}\left(x,\frac{c^2}{2}\right)$ and $h:=\text{HypergeometricPFQ}\left[\left\{\frac{3}{2},2\right\},\left\{\frac{5}{4},\frac{7}{4}\right\},-c^2\right]$. Then (according to mathematica)

$$\int\limits_0^\infty\frac{x^2}{(x+c)^{3/2}}e^{-x^2}=\\ \frac{e^{-\frac{c^2}{2}} \left(3 \pi \left(c^2 \left(-3+4 c^2\right) b[-1/4]+\left(1-7 c^2+4 c^4\right) b[1/4]-c^2 \left(-1+4 c^2\right) \left(b[3/4]+b[5/4]\right)\right)+64 c^2 e^{\frac{c^2}{2}} h\right)}{12 \sqrt{c}} $$

I am absolutely sure this does not help you at all. I am very sorry to disappoint you, but I dont know how this can be done. I know improper Integrals are usually solved by integrating over semi-circles in complex space, however that is not easily done for this function.

share|cite|improve this answer
No problem, thanks for trying. It's still helpful to know that there isn't an easy solution; now I can stop beating my head against the wall about it. – Sam Jan 30 '13 at 4:46

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.