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Math Help - 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:

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

    =  \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; August 1st 2011 at 11:15 AM.
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  2. #2
    Senior Member
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    Clarksville, ARk
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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  \eta y^2+2y\,.
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