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 $y > 0$
Differentiating under the integral sign problem
5
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calculus
derivatives
improper-integrals
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0Presumably you need $y>0$? – 2012-12-28
2 Answers
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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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0and then do what? – 2012-12-28
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0We 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. – 2012-12-28
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0but shouldn't it be $dx=du/2√y$ ? – 2012-12-28
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0No, 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 $dx=\frac{du}{\sqrt{y}}$. – 2012-12-28
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0yes right thanks a lot my friend :) – 2012-12-28
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$$u^2=x^2y\Longrightarrow2udu=2xydx\Longrightarrow dx=\frac{u}{\frac{u}{\sqrt y}y}du=\frac{du}{\sqrt y}\Longrightarrow$$
$$\int_0^\infty e^{-x^2y+1}dx=\frac{e}{\sqrt y}\int_0^\infty e^{-u^2}du=\frac{e}{2}\sqrt{\frac{\pi}{u}}$$