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E^x^2 definite integral

  1. Jan 12, 2014 #1
    I attached the solution from the solution manual of the integral I'm trying to figure out.

    [tex]\int_{-∞}^{∞}x^{2}exp(\frac{-2amx^{2}}{h})[/tex]

    The solution of that integral without the x2 in front is [itex]\sqrt{\frac{{\pi}h}{2am}}[/itex]

    So with the x2 I assumed I needed to do integration by parts.

    So taking u = x2, du = 2xdx
    And taking dv to be [itex]exp(\frac{-2amx^{2}}{h})[/itex]

    v = [itex]\sqrt{\frac{{\pi}h}{2am}}[/itex]

    But v would only equal that in a definite integral. When I'm doing integration by parts, I have an indefinite integral. So I'm kinda stuck here. When I put it into wolfram alpha, I get an error function for the answer to that indefinite integral. Do I put that answer in as v?

    Thanks.
     

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  3. Jan 12, 2014 #2

    CAF123

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    Hi leroyjenkens,
    Reexpress $$\int_{-\infty}^{\infty} x^2 \exp(-2amx^2/h) dx = \int_{-\infty}^{\infty} x \cdot x \exp(-2amx^2/h)dx$$ Does this hint at a different choice of u and v'?
     
  4. Jan 12, 2014 #3

    Ray Vickson

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    You have the wrong u and v, since your dv is not integrable in terms of elementary functions. Try again.
     
  5. Jan 12, 2014 #4
    Thanks guys.
    Yes, that was clever.
     
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