Integrating the Error-function: Using Differentiation under the Integral Sign

[Nikitin]

Homework Statement

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

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?

 

[Mark44]

Homework Statement

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

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?

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

 

[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 $\alpha=1$.

 

[brmath]

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 $\alpha=1$.

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

 

[Dick]

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

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

 

[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 :)

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 $\alpha=1$.

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?

 

[Dick]

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?

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?

 

[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?

 

[Dick]

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?

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.

 

[brmath]

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?

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 $\sqrt {\pi /\alpha}$.

 

[Nikitin]

Okay, now I fully understand. I didn't think about putting v'(x) = x e^-x^2, as I wasn't aware of

You don't apply the limits till the end. And that expression doesn't have an elementary indefinite integral.

Now everything is clear. Thanks guys!

 

[Chestermiller]

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.