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Can anyone here just tell me this is true? I just need a YES/NO, because I am a bit confused right now...

\begin{align} \int\limits_{-\infty}^{\infty}\exp\left[{-\frac{x^2}{a}}\right]dx = \left.\left( -\frac{a}{2x} \right)\exp\left[{-\frac{x^2}{a}}\right]\right|_{-\infty}^{\infty} \end{align}

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No. The right hand is not the primitive of the integrand in the left hand, if that's what you meant to write. – DonAntonio Aug 1 '13 at 11:32
Yes that is what i wanted to write. Where is the catch here??? – 71GA Aug 1 '13 at 11:33
It seems like you are trying to use the chain rule in reverse... the problem is that it doesn't work like that. You would need to use substitution; unfortunately, though, you don't have an $x$ to cancel out the derivative of $-\frac{x^2}{a}$, and so you can't carry out substitution here. – Nick Peterson Aug 1 '13 at 11:37
Have you tried differentiating the expression on the right-hand side? – Mark Bennet Aug 1 '13 at 11:37
i think you should use polar coordinates – what'sup Aug 1 '13 at 11:38
up vote 1 down vote accepted

Here's a way to find out the simplest case (understand and explain each step):

$$I:=\int\limits_{-\infty}^\infty e^{-x^2}dx\implies I^2=\left(\int\limits_{-\infty}^\infty e^{-x^2}dx\right)^2=\int\limits_{-\infty}^\infty e^{-x^2}dx\int\limits_{-\infty}^\infty e^{-y^2}dy=$$

$$=\int\limits_{-\infty}^\infty\int\limits_{-\infty}^\infty e^{-(x^2+y^2)}dxdy\stackrel{\text{polar coord.}}=\int\limits_0^\infty\int\limits_0^{2\pi}re^{-r^2}d\theta dr=$$

$$=\left.-\pi\int\limits_0^\infty(-2r\,dr)e^{-r^2}=-\pi e^{-r^2}\right|_0^\infty=-\pi(0-1)=\pi$$

and from here


Now your integral, assuming $\,a>0\,$:

$$J:=\int\limits_{-\infty}^\infty e^{-x^2/a}dx\;\ldots\;\;\text{substitution}:\;\;u:=\frac x{\sqrt a}\;,\;dx=\sqrt a\,du\implies$$

$$J=\sqrt a\int\limits_{-\infty}^\infty e^{-u^2}du=\sqrt{a\pi}$$

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oh man i just posted an answer like your answer – what'sup Aug 1 '13 at 11:46
i didn't see your answer i swear – what'sup Aug 1 '13 at 11:47
@what'sup , don't worry: if it bothers you a lot delete your answer, or else leave it as it is and let others decide which approach they like better. This happens a lot with these basic questions and no need to feel bad. – DonAntonio Aug 1 '13 at 11:48
ok thank you DonAntonio – what'sup Aug 1 '13 at 11:50
Thanks @SamiBenRomdhane , fixed. – DonAntonio Aug 1 '13 at 11:56


To find the value of the integral:

Multiply the integral by $\int\limits_{-\infty}^{\infty}\exp\left[{-\frac{y^2}{a}}\right]dy$ then use the polar coordinates.

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Nice suggestion, Sami. – amWhy Apr 25 '14 at 12:03

evaluate $$ \large{ \int_{-\infty}^{\infty} e^{\frac{-x^2}{a}} \ dx} $$

now we have $$ \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\frac{x^2+y^2}{a}} \ dx \ dy = 4\int_0^{\infty} \int_0^{\infty} e^{-\frac{x^2+y^2}{a}} \ dx \ dy $$ (ok i like 0 to inf ) to polar coordinate

$$4\int_0^{\frac{\pi}{2}} \int_0^{\infty} re^{-\frac{r^2}{a}} \ dr \ d\theta$$

now it became easy and note that

$$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\frac{x^2+y^2}{a}} \ dx \ dy = \left(\large{ \int_{-\infty}^{\infty} e^{\frac{-x^2}{a}} \ dx} \right)^2 $$

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