# Integral Problem involving e

1. May 15, 2012

### kabailey

W=∫0PdV

P=e-v2

Do I simply substitue P in the original equation, differentiate, then integrate?

W=∫0e-v2dV

W=∫0(e-v2)/2v

So far is this correct?

2. May 15, 2012

### LCKurtz

No, it isn't. That integral is usually worked by a trick of multiplying it by itself and changing it to polar coordinates. See
http://en.wikipedia.org/wiki/Gaussian_integral
There they do it from $-\infty$ to $\infty$, but the same idea applies.

3. May 15, 2012

### kabailey

I was not familiar with the Gaussian intergral. Well since it's only half the distance I would divide by 2 ∴

W=∫0e-v2dV

=(√∏)/2|0

so does that mean that the interval of ∞ to 0 is negligible???

4. May 15, 2012

### LCKurtz

The integrand is an even function so, yes, you can just divide the full integral by 2. The integal from on $[-\infty,0]$ equals the integral over $[0,\infty]$.

5. May 15, 2012