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anonymous

  • 5 years ago

solve this integral, find a in terms of b

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  1. anonymous
    • 5 years ago
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    \[\int\limits ax^2e^{-bx^2}dx\]

  2. anonymous
    • 5 years ago
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    Move the 'a' to the outside, and do some integration by parts; set u=x^2.

  3. anonymous
    • 5 years ago
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    yeah, I solved it but I used a gamma function I was wondering if it can be solved without the use of a gamma function?

  4. anonymous
    • 5 years ago
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    Sorry, I totally blanked and forgot about the whole "non-integrability" of the second term. :P I don't know how to solve this, but at a cursory glance the gamma function would seem like the most straightforward way.

  5. anonymous
    • 5 years ago
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    As far as I know of thats the only way I can solve this, and it takes really long, which sucks bc I'll have 3 more integrals like this on my exam on monday, not to mention the other 7 problems as well, that I have to complete in 50 min.

  6. anonymous
    • 5 years ago
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    Here's what Wolfram says; but it doesn't give steps. http://www.wolframalpha.com/input/?i=integrate+a*x^2*e^%28-bx^2%29

  7. anonymous
    • 5 years ago
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    This is how i solved it: \[1=\int\limits_{0}^{\infty} ax^2e^{-bx^2}dx \rightarrow bx^2=t, 2bxdx=dt\] \[\frac{a}{2b^{\frac{3}{2}}} \int\limits_{0}^{\infty} t^{\frac{1}{2}}e^{-t}dt=1\] \[\frac{a}{2b^{\frac{3}{2}}} \int\limits_{0}^{\infty} t^{\frac{3}{2}-1}e^{-t}dt=1\] since:\[\int\limits_{0}^{\infty} e^{-x}x^{\alpha-1}dx=\sqrt{\alpha}\] therefore.... \[\frac{a}{2b^{\frac{3}{2}}}\Gamma(\frac{3}{2})=1\rightarrow \frac{a}{2b^{\frac{3}{2}}}(\frac{1}{2})\sqrt{\frac{1}{2}}=1\] since: \[\sqrt{\frac{1}{2}}=\sqrt{\pi}\rightarrow \frac{a \sqrt{\pi}}{4b^\frac{3}{2}}=1\rightarrow a=\frac{4b^\frac{3}{2}}{\sqrt{\pi}} \]

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