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I tried integrating by parts but I always arrive at an expression other than $\frac{1}{2a}$ which contains $\sqrt{\frac{\pi}{a}}$ from the Gaussian integral $\int_0^\infty e^{-ax^2}dx=\frac12 \sqrt{\frac{\pi}{a}}$. Is there some kind of trick to evaluating this integral?

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  • $\begingroup$ Let $-ax^2=u$.. $\endgroup$
    – Nosrati
    May 3, 2017 at 19:08

4 Answers 4

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Hint: let $u=ax^2$. Then $1/2du=axdx$.

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  • $\begingroup$ $1/2du=axdx$ I think. $\endgroup$
    – Nosrati
    May 3, 2017 at 19:13
  • $\begingroup$ @MyGlasses, thanks, fixed. $\endgroup$
    – Paul
    May 3, 2017 at 19:14
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$\int_0^{+\infty}xe^{-ax^2}dx=\frac{1}{-2a}\int_0^{+\infty}-2axe^{-ax^2}dx=\frac{1}{-2a}[e^{-ax^2}]_0^{+\infty}=\frac{1}{2a}$

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Substitute $x^2=y$. Then $$\int_0^\infty xe^{-ax^2}\,dx=\frac12\int_0^\infty e^{-ay}\,dy.$$

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This integral is very simple and can be explained and understood in a better way just using simple properties of Gamma function Put $$ax^2 =z$$ then use the definition of Gamma function also $$\Gamma \frac{1}{2}=\sqrt{\pi}$$

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  • $\begingroup$ No need to switch to polar coordinates, switch only if you know the topic with complete understanding $\endgroup$
    – Angad
    May 3, 2017 at 19:09
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    $\begingroup$ This has nothing to do with the problem being asked. $\endgroup$
    – Paul
    May 3, 2017 at 19:10
  • $\begingroup$ I am just giving him information so that he could solve the question on his own $\endgroup$
    – Angad
    May 3, 2017 at 19:11
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    $\begingroup$ But your information doesn't help. You didn't read the problem carefully. $\endgroup$
    – Paul
    May 3, 2017 at 19:12
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    $\begingroup$ Gamma function is not required. This is a simple substitution integral. $\endgroup$
    – user307169
    May 3, 2017 at 19:20

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