9
$\begingroup$

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.

$\endgroup$
6
$\begingroup$

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 $\displaystyle \frac{\gamma}{\sqrt{\delta}}$.

$\endgroup$
6
$\begingroup$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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