Integrating exp(x^2): Solution & Explanation

[Wishe Deom]

Hello everyone,

This is my first post on the forum; I'm pretty sure my question fits in this section.

I am having a lot of difficulty finding the definite integral
\int^{+\infty}_{-\infty}e^{-2ax^{2}} dx
where a is positive and real.
I know the answer is \\sqrt{\frac{\pi}{2a}}, but I have no idea at all how to get there.

 

[fluxions]

hint: write another integral of that form, but with y as the variable. multiply them together. convert to polar coordinates, and integrate. then take the square root.

 

[Wishe Deom]

I will try this, but why would this work? As far as I understand it, the product of the integrals is not necessarily equal to the integral of the products.

Is this what you are suggesting I try?

\int{\inf}_{-\inf}e{-ax{2}}e{-ay{2}}dxdy

 

[fluxions]

\int \int f(x) g(y) dx dy = \int f(x) dx \int g(y) dy

(most of the time)

and yes, I'm suggesting you look at:

\sqrt{\int \int e^{-\alpha x^2} e^{-\alpha y^2} dxdy

 

[fluxions]

As a bonus, once you convince yourself that
\int_{-\infty}^{\infty} e^{-\alpha x^2} dx = \sqrt{{\frac{\pi}{\alpha}},
you can easily calculate integrals of the form
\int_{-\infty}^{\infty}x^{2n} e^{-\alpha x^2} dx
by differentiating the first expression (under the integral sign, on the left) with respect to alpha n times. This is a very useful fact.

 

[snipez90]

I will try this, but why would this work? As far as I understand it, the product of the integrals is not necessarily equal to the integral of the products.

To justify the interchange you can use "[URL Theorem[/URL].

 

[Wishe Deom]

Thank you for all the help. I'm well on my way, but I am having trouble with the limits of integration. Since x and y go from -inf to inf, what are the new limits?

Since r = sqrt(x^2+y^2), it seems that r goes from inf to inf, which would result in 0. I know, logically, r should be 0 to infinity, but how do I calculate that?

 

[HallsofIvy]

I will try this, but why would this work? As far as I understand it, the product of the integrals is not necessarily equal to the integral of the products.

Is this what you are suggesting I try?

\int{\inf}_{-\inf}e{-ax{2}}e{-ay{2}}dxdy

<br /> <br /> <blockquote data-attributes="" data-quote="fluxions" data-source="post: 2888425" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> fluxions said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> \int \int f(x) g(y) dx dy = \int f(x) dx \int g(y) dy <br /> <br /> (most of the time) </div> </div> </blockquote> This is &quot;Fubini&#039;s Theorem&quot;

 

[HallsofIvy]

1) \int_{-\infty}^\infty e^{-2ax^2}dx= 2\int_0^\infty e^{-2ax^2}dx [/itex]<br /> by symmetry.<br /> <br /> 2) If you let I= \int_{-\infty}^\infty} e^{-2ax^2}dx<br /> then \frac{I}{2}= \int_0^\infty e^{-2ax^2}dx<br /> and it is also true that <br /> \frac{I}{2}= \int_0^\infty e^{-2ay^2}dy<br /> since that is the same integral with a different &quot;dummy&quot; variable.<br /> <br /> so<br /> 3) I^2/4= \left(\int_0^\infty e^{-2ax^2}dx\right)\left(\int_0^\infty e^{-2ay^2}dy\right)<br /> = \int_0^\infty \int_0^\infty e^{-2a(x^2+ y^2)} dy dx<br /> by Fubini&#039;s theorem.<br /> <br /> You can interpret that last as a double integral over the first quadrant and change to polar coordinates. Remember that the &quot;differential of area&quot; in polar coordinates is r drd\theta and that \theta goes from 0 to \pi/2 in the first quadrant.