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Math Help - an integration problem with exponantial

  1. #1
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    an integration problem with exponantial

    I have encountered a problem with the following form:

    \int x e^{ax^2-bx}dx

    I can only find formulas for \int x e^{ax^2}dx and \int x e^{bx}dx and don't know how to make use of these two formulas.

    Can someone help me on this? Thanks a lot!!
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  2. #2
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    Quote Originally Posted by billzhao View Post
    I have encountered a problem with the following form:

    \int x e^{ax^2-bx}dx

    I can only find formulas for \int x e^{ax^2}dx and \int x e^{bx}dx and don't know how to make use of these two formulas.

    Can someone help me on this? Thanks a lot!!
    Complete the square in the exponent to get

    \displaysytle a(x^2-\frac{b}{a}x+\frac{b^2}{4a^2})-\frac{b^2}{4a}=a(x-\frac{b}{a})^2-\frac{b^2}{4a}

    Now the integral looks like

    \displaystyle \int xe^{a(x-\frac{b}{2a})^2-\frac{b^2}{4a}}dx=e^{-\frac{b^2}{4a}}\int xe^{a(x-\frac{b}{2a})^2} dx

    From here just make a u sub for the exponent and then you will use both of your formula's for the integral
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  3. #3
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    Thanks for the quick reply!! But...

    Is there a closed form solution for the below integration?

    \int e^{ax^2}dx
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  4. #4
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    Yes, but you have to use the "special function" erf(x)= \frac{2}{\sqrt{\pi}}\int_0^x e^{-t^2}dt.

    Let iu= \sqrt{a}x. Then dx= \frac{i}{\sqrt{a}}du, e^{ax^2}= e^{-u^2} and the integral becomes \frac{i}{\sqrt{a}}\int e^{-u^2}du= \frac{i\sqrt{\pi}}{2\sqrt{a}}erf(-i\sqrt{a}x)+ C.

    The fact that you have x^2 in the exponential rather than -x^2 complicates things and causes those "i" factors.
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