# Error-function integral

1. Oct 14, 2013

### Nikitin

1. The problem statement, all variables and given/known data
If you know
$$\int_{-\infty}^{\infty} e^{-\alpha \beta^2} d \beta = \sqrt{\pi / \alpha}$$
Show that
$$\int_{-\infty}^{\infty} \beta^2 e^{-\beta^2} d \beta = \sqrt{\pi}/2$$

3. The attempt at a solution

I have tried integration by parts on the second integral, but to no avail. I just end up with it being equal to zero. What am I to do?

2. Oct 14, 2013

### Staff: Mentor

How did you split the integrand up when you did integration by parts?

3. Oct 14, 2013

### Dick

You don't really need to do integration by parts. Take the derivative with respect to $\alpha$ of both sides of your first equation. Then set $\alpha=1$.

4. Oct 14, 2013

### brmath

This is a very nice trick, and I like it a lot. But it could be Nikitin needs to work on integration by parts.

5. Oct 14, 2013

### Dick

Yes, probably. It's easy enough to do that way too.

6. Oct 15, 2013

### Nikitin

Yeah, I must've done a silly mistake when doing the integration by parts. I set the $\beta^2$ term as u, and the $e^{ -\beta^2}$ term as v'.

Then, (btw, how do I put limits on the left part of a bracket in latex?)

$$\int_{-\infty}^{\infty} \beta^2 e^{-\beta^2} d \beta = \left [ \beta^2 \int e^{-\beta^2} d \beta \right ]_{-\infty}^{\infty} - \int_{-\infty}^{\infty} 2 \beta (\int_{-\infty}^{\infty} e^{-\beta^2} d \beta ) d \beta$$

If I try to insert
$$\int_{-\infty}^{\infty} e^{-\beta^2} d \beta = \sqrt{\pi}$$

into my integration by parts expression, I will just end up with 0. Can somebody please do it properly so I can find my mistake?

PS: Sorry for late reply. Oh, and thanks for all the help and nice tips :)

I don't want to be a bore (I like your method), but how do you know it's possible to differentiate the term inside the integral first, instead of taking the integral first and then differentiate?

Last edited: Oct 15, 2013
7. Oct 15, 2013

### Dick

http://en.wikipedia.org/wiki/Leibniz_integral_rule And for your integration by parts, put $dv=\beta e^{-\beta^2} d\beta$. What does that leave for u?

Last edited: Oct 15, 2013
8. Oct 15, 2013

### Nikitin

I don't completely understand why you'd do that?

Is the integration-by-parts expression in post #6 correct? Why can't I just insert $\int_{-\infty}^{\infty} e^{-\beta^2} d \beta = \sqrt{\pi}$ into the expression and solve?

9. Oct 15, 2013

### Dick

If you are going to set $dv=e^{\beta^2}d\beta$ then v is the INDEFINITE integral of dv. You don't apply the limits till the end. And that expression doesn't have an elementary indefinite integral. Hence my alternate suggestion.

10. Oct 15, 2013

### brmath

Re taking the derivative under the integral, note that the integral is with respect to $\beta$. For purposes of regarding $\alpha$ as the variable to differentiate against, all the $\beta$ stuff is essentially a constant.

Re integration by parts take u(x) = x and v'(x) = x$e^{-\alpha x^2}$. The new integral will pop up looking very like the one you know to be $\sqrt {\pi /\alpha}$.

11. Oct 15, 2013

### Nikitin

Okay, now I fully understand. I didn't think about putting v'(x) = x e^-x^2, as I wasn't aware of
Now everything is clear. Thanks guys!

12. Oct 15, 2013

### Staff: Mentor

Getting back to Dick's method in response #3, I think what your textbook is trying to teach you is the technique of "differentiation under the integral sign." I think this is how the book intended for you to solve the problem.