Can you do this Polar Coordinate Double Integral Problem

The question I'm doing is 4 parts, but the only part I need help on is proving that the integral from -infinity to +infinity of e^(-x^2) dx = (pi)^1/2.

The previous parts we turned e^-(x^(2)+y^(2)) from a box with +/- infinity to a disk with radius a (which goes to infinity) and ranges from 0 to 2pi.

For the part I need help on, I tried the double integral of e^(-r^(2)(cos(theta))^2) but I haven't gotten the answer I'm looking for, and I can't integrate e^((cos(theta))^2). and I don't even know if I'm supposed to use a double integral for this part, but I'm pretty sure I am.

Can someone please help me?

Let I = ∫(-∞ to ∞) e^(-x^2) dx.

So, I^2 = ∫(-∞ to ∞) e^(-x^2) dx * ∫(-∞ to ∞) e^(-y^2) dy
...........= ∫(-∞ to ∞) ∫(-∞ to ∞) e^(-x^2-y^2) dy dx.

Converting to polar coordinates yields
I^2 = ∫(θ = 0 to 2π) ∫(r = 0 to ∞) e^(-r^2) * (r dr dθ); note that there are no cosines in sight.
.....= ∫(θ = 0 to 2π) dθ ∫(r = 0 to ∞) r e^(-r^2) dr
.....= 2π * (-1/2)e^(-r^2) {for r = 0 to ∞}
.....= π.

Since I > 0, taking square roots yields I = √π.

I hope this helps!

Yes that is. The fact that the bounds are constant helps, too, in that the order of integration is irrelevant (convergence issues aside, which don't matter in this problem).

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