Polar Coordinates Improper Integral Proofs

[themadhatter1]

Homework Statement

(a) we define the improper integral (over the entire plane R²)
I=\int\int_{R^2}e^{-(x^2+y^2)}dA=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}dy dx=\lim_{a\rightarrow\infty}\int\int_{D_{a}} e^{-(x^2+y^2)} dA

where D_a is the disk with radius a and center the origin. Show that
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}dA=\pi

(b)
An equilivent definition of the improper integral in part (a) is
\int\int_{R^2}e^{-(x^2+y^2)}dA=\lim_{a\rightarrow\infty}\int\int_{S_{a}} e^{-(x^2+y^2)} dA

where S_a is the square with vertices (\pm a,\pm a) Use this to show that
\int_{-\infty}^{\infty}e^{-x^2}dx\int_{-\infty}^{\infty}e^{-y^2}dy=\pi

(c) deduce that

\int_{-\infty}^{\infty}e^{-x^2}dx=\sqrt{\pi}

(d) By making the change of variable t=\sqrt{2}x. show that
\int_{-\infty}^{\infty}e^{\frac{-x^2}{2}}dx=\sqrt{2\pi}

Homework Equations

The Attempt at a Solution

I proved a by doing a change of variables into polar coordinates. However, I'm not quite sure how I would go about proving b.The idea that S_a is a square means I would be using Cartesian coordinates however, those 2 integrals don't have an antiderviative, what should I look at next?

 

[themadhatter1]

update, I proved everything, however I'm not sure If my proof for b is what they're asking for. I said that

\int_{-\infty}^{\infty}e^{-x^2}dx\int_{-\infty}^{\infty}e^{-y^2}dy=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}dy dx

and since I had proved that the rhs of the equation equals pi it must be true that the left hand side of the equation equals pi as well. doesn't a square of infinite dimensions have the same area as a polar rectangle where 0≤θ≤2∏ and 0≤r≤infinity. Does this satisfy the question being asked?

 

[AnthonyLiardo]

That's exactly what I did for part b. I have this assigned so I need some help on this problem as well.

I don't know how to do part c yet. I took the equation from b, and divided by the e^(-(y^2)) integral and all I would have to do is prove that that integral equals √∏. I tried to say that -y^2=x^2-r^2 so if you replace the e^(-(y^2)) integral with e^(x^2)/e^(r^2) (which I'm not sure if you can do), then idk I'm kinda stuck.