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I have...

$$ I = \int_{\sqrt{R^{2}-\left(R-\epsilon\right)^{2}}}^{R}dxe^{-\beta x^{2}}Erf\left[\sqrt{\beta}\left(R-\epsilon-\sqrt{R^{2}-x^{2}}\right)\right]\ $$

(everything's a constant except $x$)

...and someone told me that differentiating $I$ could help evaluate it.

Can someone explain? Or help evaluate this? I've never really seen this kind of integral before, but I heard it's do-able.

Cheers!

apg
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  • Have a look to Sasha's answer in this post: http://math.stackexchange.com/questions/78615/integrating-a-product-of-exponentials-and-error-functions?rq=1. Maybe you can use it to solve your problem. – Dmoreno Jan 19 '14 at 23:20
  • has this guy just differentiated the integrand in the first answer? – apg Jan 20 '14 at 00:01
  • Yes, but he differentiated with respect to a parameter in order to get an easier integral. In your case, this parameter could be $\epsilon$, $R$ or $\beta$ but in every case this doesn't make things much more easier. A faster way to compute $I$ would definitely be numerical integration. Cheers! – Dmoreno Jan 20 '14 at 00:06

1 Answers1

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The integrand seems not to depend on $y$. Why is that?

Anyway, I'll give you a hint:

$$ I = \int \text{Erf}(x) e^{-{x^2}} dx = \frac{\sqrt{\pi}}{2} \text{Erf}^2(x) - \frac{\sqrt{\pi}}{2} \int \frac{2}{\sqrt{\pi}} e^{-x^2} \text{Erf}(x) dx = \frac{\sqrt{\pi}}{2} \text{Erf}^2(x) - I,$$

then:

$$ I = \frac{\sqrt{\pi}}{4} \text{Erf}^2(x).$$

I hope this may help you.

Cheers!

To evaluate the integral I often use this "trick" (chain rule):

$$\int f g \, dx = F g - \int F g' \, dx,$$

where $F = \int f \, dx$ is a primitive of $f$. Subsitute $f = e^{-x^2}$ and $g = \text{Erf}(x)$ and take into account that: $\int e^{-x^2} \, dx = \sqrt{\pi} \, \text{Erf}(x)/2$.

Dmoreno
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  • sorry it should just be dx not dxdy. – apg Jan 19 '14 at 23:54
  • so basically, just use your trick for integrals of this form? No need for a substitution? – apg Jan 20 '14 at 00:05
  • Notice that the argument of the error function in your case is much more complicated so my "trick" is not suitable for it to be used. Maybe thinking about a change of variable could help, but I can't already see it. – Dmoreno Jan 20 '14 at 00:08