Evaluate the integral from standard from

[osker246]

Homework Statement

evaluate [itex]\int[/itex]e^([itex]\frac{\kappa*x^2}{2KT}[/itex])dx with limits of integration from -infinity to +infinity using the standard form [itex]\int[/itex]e^(-C*x²)dx = ([itex]\frac{\pi}{4C}[/itex])^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).

The Attempt at a Solution

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

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

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

 

[Unit]

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

 

[osker246]

Thanks for the reply. I did see that it was an even function, that's why I changed my limits of intergration to 0 to +infinity and multiply the answer by two. But the integral of the standard function doesn't keep X as a variable. So how do I evaluate this integral with my limits of intergration? I feel like I am over looking something here.

 

[dextercioby]

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

 

[Unit]

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

 

[osker246]

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