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Thread: Long integral, not sure how the algebra works in there

  1. #1
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    Long integral, not sure how the algebra works in there

    In my textbook, it reads:

    $\displaystyle 2 \int ^ { \infty } _0 e ^ { - \eta y^2-2y } dy $

    =$\displaystyle \frac { 2 \sqrt { \pi } e^ { \frac {1} { \eta } }} { \sqrt { \eta }}} \int ^ \infty _0 \frac { \exp [ \frac { - ( \frac {y+1}{ \eta })^2 }{ \sqrt { \frac {2 }{2 \eta }}} ] }{ \sqrt { \frac {2 \pi }{2 \eta } } } dy $

    Why?
    Last edited by tttcomrader; Aug 1st 2011 at 11:15 AM.
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  2. #2
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    Re: Long integral, not sure how the algebra works in there

    It looks like the author completed the square for the exponent of e.

    In other words, complete the square for $\displaystyle \eta y^2+2y\,.$
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