Evaluate the integral from standard from

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SUMMARY

The integral of the function \(\int e^{\frac{\kappa x^2}{2KT}} dx\) from \(-\infty\) to \(+\infty\) can be evaluated using the standard form \(\int e^{-Cx^2} dx = \left(\frac{\pi}{4C}\right)^{1/2}\) with \(C = \frac{-\kappa}{2KT}\). The correct evaluation leads to the result \(2\left(\frac{-2KT\pi}{4\kappa}\right)^{1/2}\). The discussion highlights the importance of recognizing the even function property of the integrand, allowing the limits of integration to be adjusted from \(-\infty\) to \(+\infty\) to \(0\) to \(+\infty\) and then multiplied by two.

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Homework Statement


evaluate \inte^(\frac{\kappa*x^2}{2KT})dx with limits of integration from -infinity to +infinity using the standard form \inte^(-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).





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?
 
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Hint: exp(-Cx2) is an even function, symmetric about the y-axis.
 
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.
 
You needn't evaluate any limit. The integral you're given already takes into account the limits at infinity and 0.
 
Ah, my mistake osker246. And yes, dextercioby is right.
 
So does this mean my answer is 2*(\frac{-2KT\pi}{4\kappa})1/2?
 

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