# Evaluate the integral from standard from

1. Jun 26, 2011

### osker246

1. The problem statement, all variables and given/known data
evaluate $\int$e^($\frac{\kappa*x^2}{2KT}$)dx with limits of integration from -infinity to +infinity using the standard form $\int$e^(-C*x2)dx = ($\frac{\pi}{4C}$)1/2 with limits of integration from 0 to +infinity. Note κ, k, and T are constants. In the standard form c indicates a constant. Note the function being integrated is an even function: f(x)=f(-x).

3. The attempt at a solution

Well looking at the equation I see C=$\frac{-\kappa}{2KT}$. I then plug C into ($\frac{\pi}{4C}$)1/2giving ($\frac{-2KT\pi}{4\kappa}$)1/2.

My next step would be to evaluate:
2*[($\frac{-2KT\pi}{4\kappa}$)1/2]$^{+infinity}_{0}$

But I no longer have my variable x to do so, am I missing something? Is my answer simply ($\frac{-2KT\pi}{4\kappa}$)1/2?

2. Jun 26, 2011

### Unit

Hint: exp(-Cx2) is an even function, symmetric about the y-axis.

3. Jun 26, 2011

### osker246

Thanks for the reply. I did see that it was an even function, thats why I changed my limits of intergration to 0 to +infinity and multiply the answer by two. But the integral of the standard function doesnt keep X as a variable. So how do I evaluate this integral with my limits of intergration? I feel like im over looking something here.

4. Jun 26, 2011

### dextercioby

You needn't evaluate any limit. The integral you're given already takes into account the limits at infinity and 0.

5. Jun 26, 2011

### Unit

Ah, my mistake osker246. And yes, dextercioby is right.

6. Jun 26, 2011

### osker246

So does this mean my answer is 2*($\frac{-2KT\pi}{4\kappa}$)1/2?