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Evaluate the integral from standard from

  1. Jun 26, 2011 #1
    1. The problem statement, all variables and given/known data
    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).





    3. 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?
     
  2. jcsd
  3. Jun 26, 2011 #2
    Hint: exp(-Cx2) is an even function, symmetric about the y-axis.
     
  4. Jun 26, 2011 #3
    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.
     
  5. Jun 26, 2011 #4

    dextercioby

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    You needn't evaluate any limit. The integral you're given already takes into account the limits at infinity and 0.
     
  6. Jun 26, 2011 #5
    Ah, my mistake osker246. And yes, dextercioby is right.
     
  7. Jun 26, 2011 #6
    So does this mean my answer is 2*([itex]\frac{-2KT\pi}{4\kappa}[/itex])1/2?
     
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