Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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}

share|improve this question
1  
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. –  Nicholas R. Peterson Aug 1 '13 at 11:37
1  
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

3 Answers 3

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

$$I=\sqrt\pi$$

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

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

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

share|improve this answer
    
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

Hint

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.

share|improve this answer
    
Nice suggestion, Sami. –  amWhy Apr 25 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

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.