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

I have the following integral $$ \int\limits_0^\infty x^2\exp(-\delta x^2)\operatorname{erf}(\gamma x)\,dx. $$
Ideally, I would like a closed-form in terms of common functions, but a series answer will do.

share|cite|improve this question

I think it's best to just proceed as is natural

$\displaystyle \begin{aligned}\int_0^{\infty}x^2e^{-\delta x^2}\text{erf}(\gamma x)\; dx &= \frac{2}{\sqrt{\pi}}\sum_{n=0}^{\infty}\frac{(-1)^n\gamma^{2n+1}}{n!(2n+1)}\int_0^{\infty} e^{-\delta x^2}x^{2n+3}\; dx\\ &= \frac{1}{\sqrt{\pi}}\sum_{n=0}^{\infty}\frac{(-1)^n\gamma^{2n+1}\Gamma(n+2)}{n!(2n+1)\delta^{n+2}}\\ &=\frac{1}{\sqrt{\pi}}\sum_{n=0}^{\infty}\frac{(-1)^n\gamma^{2n+1}(n+1)}{\delta^{n+2}(2n+1)}\\ &=\frac{1}{\delta^{\frac{3}{2}}\sqrt{\pi}}\left(\frac{1}{2}\sum_{n=0}^{\infty}(-1)^n\left(\frac{\gamma}{\sqrt{\delta}}\right)^{2n+1}+\frac{1}{2}\sum_{n=0}^{\infty}\frac{(-1)^n\left(\frac{\gamma}{\sqrt{\delta}}\right)^{2n+1}}{2n+1}\right)\\ &=\frac{1}{2\delta^{\frac{3}{2}}\sqrt{\pi}}\left(\frac{\gamma\sqrt{\delta}}{\gamma^2+\delta}+\arctan\left(\frac{\gamma}{\sqrt{\delta}}\right)\right)\end{aligned}$

Where the only non-trivial facts used was the common expression $\displaystyle \int_0^{\infty}e^{-x^2}x^n\; dx=\frac{\Gamma\left(\frac{n+1}{2}\right)}{2}$. Also, note that I used the Maclaurin series for arctangent, and so we need to have a restriction on $\displaystyle \frac{\gamma}{\sqrt{\delta}}$.

share|cite|improve this answer

Let $\mathcal{I}(\gamma)= \int_0^\infty x^2\exp(-\delta x^2)\operatorname{erf}(\gamma x)\, \mathrm{d} x$, then $I^\prime(\gamma) = \frac{2}{\sqrt{\pi}} \int_0^\infty x^3 \exp(-x^2 \left( \delta + \gamma^2 \right) ) \mathrm{d} x = \frac{1}{(\gamma^2 + \delta)^2} \frac{1}{\sqrt{\pi}}$.

Integrating: $$ I(\gamma) = \int_0^\gamma \frac{1}{\sqrt{\pi}} \frac{1}{(\gamma^2+\delta)^2} \mathrm{d} \gamma = \frac{\gamma }{2 \sqrt{\pi } \delta \left(\gamma ^2+\delta \right)}+\frac{1}{2 \sqrt{\pi } \delta ^{3/2}} \arctan\left(\frac{\gamma }{\sqrt{\delta }}\right) $$

share|cite|improve this answer
1  
Ah, I like it. Error function and exponential should have screamed differentiation inside of the integral. Nice one! – Alex Youcis Nov 3 '11 at 18:08
    
can you explain why you've integrated between 0 and gamma in the final step? – Alexander Giles Feb 14 '14 at 20:25
1  
@AlexanderGiles Because, assuming $f(x)$ is differentiable $f(x) = f(x) + \int_0^x f^\prime(z) \mathrm{d}z$. – Sasha Feb 15 '14 at 5:32
1  
you mean $f(0)$ instead of $f(x)$ on the RHS? – Alexander Giles Feb 15 '14 at 15:30
    
so essentially this is because your first line in the answer $I(\gamma)$ is $0$ at $\gamma=0$? – Alexander Giles Feb 15 '14 at 15:36

Your Answer

 
discard

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.