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Knowing that $\int_0^\infty e^{-x^2}\,dx = \frac{\sqrt{\pi}}{2},$ evaluate the integral $\int_0^\infty e^{-x^2y+1}\,dx.$ for $y > 0$

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    Presumably you need y>0?2012-12-28

2 Answers 2

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Hint: The integral diverges if $y \le 0$. For $y\gt 0$, let $x\sqrt{y}=u$.

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    yes right thanks a lot my friend :)2012-12-28
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$u^2=x^2y\Longrightarrow2udu=2xydx\Longrightarrow dx=\frac{u}{\frac{u}{\sqrt y}y}du=\frac{du}{\sqrt y}\Longrightarrow$

$\int_0^\infty e^{-x^2y+1}dx=\frac{e}{\sqrt y}\int_0^\infty e^{-u^2}du=\frac{e}{2}\sqrt{\frac{\pi}{u}}$