1. Double integrals

Find a solution to this integral $I = \int_0^{\infty} e^{-x^2} dx$ analytically by calculating $I^2 = \int_0^{\infty} \int_0^{\infty} e^{-x^2} e^{-y^2} dydx$.

I am not sure about how to go about this question, how did they get $I^2 = \int_0^{\infty} \int_0^{\infty} e^{-x^2} e^{-y^2} dydx$? Shouldn't $I^2 = \int_0^{\infty} e^{-x^2} dx \times \int_0^{\infty} e^{-x^2} dx = \int_0^{\infty}\int_0^{\infty} e^{-x^2} \cdot e^{-x^2} dxdy$?

Thanks very much! (Just a bit confused on this question because I am working ahead of class )

2. Originally Posted by usagi_killer
Find a solution to this integral $I = \int_0^{\infty} e^{-x^2} dx$ analytically by calculating $I^2 = \int_0^{\infty} \int_0^{\infty} e^{-x^2} e^{-y^2} dydx$.

I am not sure about how to go about this question, how did they get $I^2 = \int_0^{\infty} \int_0^{\infty} e^{-x^2} e^{-y^2} dydx$? Shouldn't $I^2 = \int_0^{\infty} e^{-x^2} dx \times \int_0^{\infty} e^{-x^2} dx = \int_0^{\infty}\int_0^{\infty} e^{-x^2} \cdot e^{-x^2} dxdy$?

Thanks very much! (Just a bit confused on this question because I am working ahead of class )

3. Originally Posted by usagi_killer
Find a solution to this integral $I = \int_0^{\infty} e^{-x^2} dx$ analytically by calculating $I^2 = \int_0^{\infty} \int_0^{\infty} e^{-x^2} e^{-y^2} dydx$.

I am not sure about how to go about this question, how did they get $I^2 = \int_0^{\infty} \int_0^{\infty} e^{-x^2} e^{-y^2} dydx$? Shouldn't $I^2 = \int_0^{\infty} e^{-x^2} dx \times \int_0^{\infty} e^{-x^2} dx = \int_0^{\infty}\int_0^{\infty} e^{-x^2} \cdot e^{-x^2} dxdy$?

Thanks very much! (Just a bit confused on this question because I am working ahead of class )
There is a typo in your final line- you surely don't want $\int_0^\infty e^{x^2} dx\times \int_0^\infty e^{-x}dx= \int_0^\infty\int_0^\infty e^{-x^2}e^{-x^2} dx dy$. Where did that final "dy" come from?

I suspect you meant $\int_0^\infty e^{x^2} dx\times \int_0^\infty e^{-x}dx= \int_0^\infty\int_0^\infty e^{-x^2}e^{-x^2} dx dx$ but that is also wrong. It is true that $I^2= \int_0^\infty e^{-x^2}dx\time\int_0^\infty e^{-x^2} dx$ but that is NOT the same as $\int_0^\infty\int_0^\infty e^{-x^2} e^{-x^2}dx dx$. In fact, that last double integral makes no sense. You cannot integrate twice with respect to the same variable.

Remember that "x" is a dummy variable- the final answer will be a number so that it doesn't matter what you call the variable.
If $I= \int_0^\infty e^{-x^2}dx$ then it is also true that $I= \int_0^\infty e^{-y^2}dy$ or, for that matter, $I= \int_0^\infty e^{-t^2}dt$ or $I= \int_0^\infty e^{-u^2}du$.

We then can say that $I^2= \left(\int_0^\infty e^{-x^2}dx\right)\left(\int_0^\infty e^{-y^2} dy\right)= \int_{y= 0}^\infty \int_{x= 0}^\infty e^{-x^2}e^{-y^2} dx dy$
That last part, saying that the product of the two integrals (each integral involving a different variable) is equal to the double integral is a version of "Fubini's theorem".

Of course, $e^{-x^2}e^{-y^2}= e^{-(x^2+ y^2)}$ so we can write
$I^2= \int_{y=0}^\infty\int_{x=0}^\infty e^{-(x^2+ y^2)} dx dy$.

The whole point of doing that is that now the integral is over the entire first quadrant and we can cast the integral into polar coordinates with $\theta$ going from 0 to $\pi/2$ and r from 0 to $\infty$. Of course, in polar coordinates, $x^2+ y^2= r^2$ and $dx dy= r dr d\theta$ so we have
$I^2= \int_{\theta= 0}^{\pi/2}\int_{r= 0}^\infty e^{-r^2} r drd\theta= \left(\int_{\theta=0}^{\pi/2} d\theta\right)\left(\int_{r=0}^\infty e^{-r^2}r dr\right)$
where I have used Fubini's theorem the other way to separate that double integral into a product of integrals.

Now that $\theta$ integral is easy and, because of the extra "r" so is the r integral.

4. Thanks so much both of you, I read the wiki article and HallsofIvy's great explanation, understand it now!