1. Mar 22, 2013

Tonks93

1. The problem statement, all variables and given/known data

I am trying to integrate $\int \frac{exp(-(1/4)x^2)}{(1+1/2(x^2))^2}\,dx$

2. Relevant equations

3. The attempt at a solution

I've tried all the substitutions I can think of and I'm not getting anywhere. I know that the answer cannot be calculated fully (contains an erf term), but I can't even get this out.

Any advice would be great, thank you :)

2. Mar 22, 2013

SammyS

Staff Emeritus
Hello Tonks93. Welcome to PF !

The substitution u = x/2 will clean this up a little.

Just to be clear, is the integral you're trying to solve
$\displaystyle \int \frac{exp(-(x^2/4))}{(1+(x^2/2))^2}\,dx \ ?$​

3. Mar 22, 2013

Tonks93

Thanks for your reply - I'll try what you said now! And yes that's the right integral - sorry it wasn't totally clear.

4. Mar 23, 2013

haruspex

u=x/√2 looks marginally better, but it's still a long way short of anything very useful.
I can't see how to make any real progress with this as an indefinite integral, but there are methods that might help if it's for a specific range like -∞ to +∞.

5. Mar 23, 2013

Tonks93

Thanks for your reply. I think the range should be 0 to +∞. Would this help at all?

6. Mar 23, 2013

SammyS

Staff Emeritus
As you mentioned in the Original Post, the indefinite integral contains a term with the error function.

It looks as though you will need to use integration by parts, so I suggest using the following substitution.

Let t = x/2 . (Using t = x/√2 is OK too.)

Your integral becomes: $\displaystyle \ \int \frac{exp(-(t^2))}{(1+2t^2)^2}\,dt \ .$

Then do integration by parts using $\displaystyle \ u=\frac{exp(-(t^2))}{t}\$ and $\displaystyle \ dv=\frac{t}{(1+2t^2)^2}\,dt\ .$

7. Mar 24, 2013

Tonks93

Thanks for your reply. I got the answer out with the substitution you said, so thankyou so much! :)

8. Mar 24, 2013

SammyS

Staff Emeritus
That was a crazy pair to use for integration by parts, wasn't it ?

9. Mar 25, 2013

Tonks93

Haha well the substitution seems fairly logical now you've pointed it out, but it's not one I'd have thought of myself! Thanks again :)