use double integrals
-
Start out by letting:
I = ∫ e^(-x^2) dx (from x=-∞ to ∞) = ∫ e^(-y^2) dy (from y=-∞ to ∞).
(Note: x and y are just dummy variables, so we can just replace x with y or any letter.)
Multiplying these two together and using properties of double integrals:
I * I = ∫ e^(-x^2) dx (from x=-∞ to ∞) * ∫ e^(-y^2) dy (from y=-∞ to ∞)
==> I^2 = ∫∫ e^[-(x^2 + y^2)] dx dy (from x=-∞ to ∞) (from y=-∞ to ∞).
By definition:
∫∫ e^[-(x^2 + y^2)] dx dy (from x=-∞ to ∞) (from y=-∞ to ∞) = ∫∫D₁ e^[-(x^2 + y^2)] dA,
where D₁ is the square [-s, s] x [-s, s] and s --> ∞.
Since s --> infinity, we can also describe D as the circle x^2 + y^2 = r^2 with r --> ∞. Thus, an alternative representation in polar coordinates is:
D₂ = {(x, y) | 0 <= θ <= 2π, 0 <= r <= s} (with s --> ∞).
So:
I^2 = ∫∫D₁ e^[-(x^2 + y^2)] dA
= ∫∫D₂ r*e^(-r^2) dA, from above
= lim (s-->∞) ∫∫ r*e^(-r^2) dr dθ (from r=0 to s) (from θ=0 to 2π)
= ∫ dθ (from θ=0 to 2π) * lim (s-->∞) ∫ r*e^(-r^2) dr (from r=0 to s)
= [θ (evaluated from θ=0 to 2π)][lim (s-->∞) (-1/2)e^(-r^2) (evaluated from r=0 to s)]
= (-1/2)(2π - 0) * lim (s-->∞) [e^(-s^2) - 1]
= (-1/2)(2π - 0)(0 - 1), since lim (s-->∞) e^(-s^2) = 0
= π.
Note that I = ∫ e^(-x^2) dx (from x=-∞ to ∞) > 0 as e^(-x^2) is strictly positive on (-∞, ∞), we can take the positive square root of both sides to get:
I = ∫ e^(-x^2) dx (from x=-∞ to ∞) = √π. Q.E.D.
I hope this helps!
I = ∫ e^(-x^2) dx (from x=-∞ to ∞) = ∫ e^(-y^2) dy (from y=-∞ to ∞).
(Note: x and y are just dummy variables, so we can just replace x with y or any letter.)
Multiplying these two together and using properties of double integrals:
I * I = ∫ e^(-x^2) dx (from x=-∞ to ∞) * ∫ e^(-y^2) dy (from y=-∞ to ∞)
==> I^2 = ∫∫ e^[-(x^2 + y^2)] dx dy (from x=-∞ to ∞) (from y=-∞ to ∞).
By definition:
∫∫ e^[-(x^2 + y^2)] dx dy (from x=-∞ to ∞) (from y=-∞ to ∞) = ∫∫D₁ e^[-(x^2 + y^2)] dA,
where D₁ is the square [-s, s] x [-s, s] and s --> ∞.
Since s --> infinity, we can also describe D as the circle x^2 + y^2 = r^2 with r --> ∞. Thus, an alternative representation in polar coordinates is:
D₂ = {(x, y) | 0 <= θ <= 2π, 0 <= r <= s} (with s --> ∞).
So:
I^2 = ∫∫D₁ e^[-(x^2 + y^2)] dA
= ∫∫D₂ r*e^(-r^2) dA, from above
= lim (s-->∞) ∫∫ r*e^(-r^2) dr dθ (from r=0 to s) (from θ=0 to 2π)
= ∫ dθ (from θ=0 to 2π) * lim (s-->∞) ∫ r*e^(-r^2) dr (from r=0 to s)
= [θ (evaluated from θ=0 to 2π)][lim (s-->∞) (-1/2)e^(-r^2) (evaluated from r=0 to s)]
= (-1/2)(2π - 0) * lim (s-->∞) [e^(-s^2) - 1]
= (-1/2)(2π - 0)(0 - 1), since lim (s-->∞) e^(-s^2) = 0
= π.
Note that I = ∫ e^(-x^2) dx (from x=-∞ to ∞) > 0 as e^(-x^2) is strictly positive on (-∞, ∞), we can take the positive square root of both sides to get:
I = ∫ e^(-x^2) dx (from x=-∞ to ∞) = √π. Q.E.D.
I hope this helps!