How can I prove that the error function is normalized? That is; erf(infinity) = 1?

By definition,

erf(x) = (2/√π) ∫(t = 0 to x) e^(-t^2) dt.

To show that this is normalized, we need to show that

(2/√π) ∫(t = 0 to ∞) e^(-t^2) dt = 1.

This is equivalent to the following:

(1/√π) ∫(-∞ to ∞) e^(-t^2) dt = 1, since the integrand is even

<==> ∫(-∞ to ∞) e^(-t^2) dt = √π.

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This latter integral is the well-known Gaussian integral, whose value can be deduced via polar coordinates:

Let I = ∫(-∞ to ∞) e^(-t^2) dt.

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

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

==> I^2 = ∫(-∞ to ∞) ∫(-∞ to ∞) e^(-(x^2 + y^2) dx dy.

Converting to polar coordinates yields

I^2 = ∫(θ = 0 to 2π) ∫(r = 0 to ∞) e^(-r^2) * (r dr dθ)

......= 2π * (-1/2)e^(-r^2) {for r = 0 to ∞}

......= π.

Since I > 0, we conclude that I = √π, as required.

I hope this helps!