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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 $y>0$? –  copper.hat Dec 28 '12 at 19:26

2 Answers 2

up vote 4 down vote accepted

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

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and then do what? –  Muhammad Khalifa Dec 28 '12 at 19:32
    
We are integrating $e\cdot e^{-x^2y}$. Let $u$ be as in post. Then $dx=\frac{du}{\sqrt{y}}$. Do the substitution. We want $\int_0^\infty \frac{e}{\sqrt{y}}e^{-u^2}\,du$. This is a "constant" times an integral you have been told the value of. –  André Nicolas Dec 28 '12 at 19:38
    
but shouldn't it be $dx=du/2√y$ ? –  Muhammad Khalifa Dec 28 '12 at 19:46
    
No, it shouldn't. We have $u=x\sqrt{y}$. Think of $y$ as a constant. Then $\frac{du}{dx}=\sqrt{y}$, so $du=\sqrt{y}\,dx$, or equivalently $dx=\frac{du}{\sqrt{y}}$. –  André Nicolas Dec 28 '12 at 19:50
    
yes right thanks a lot my friend :) –  Muhammad Khalifa Dec 28 '12 at 19:56

$$u^2=x^2y\Longrightarrow2udu=2xydx\Longrightarrow dx=\frac{u}{\frac{u}{\sqrt y}y}du=\frac{du}{\sqrt y}\Longrightarrow$$

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

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