Long integral, not sure how the algebra works in there

**tttcomrader** · July 31st 2011, 12:42 PM

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?

**SammyS** · July 31st 2011, 04:26 PM

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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