How would I calculate the definite integral of e^(-k/2*x^2) dx over the interval of (0,+∞)
I know the answer is sqrt(pi/(2k)), but how would I get that?
You can substitute x = sqrt(2/k) y. That yields:
I = integral over (0,+∞)
of e^(-k/2*x^2) dx =
sqrt(2/k) integral over (0,+∞)
of e^(-y^2) dy
Then, the result follows from the fact that
integral over (0,+∞) of e^(-y^2) dy =
1/2 sqrt(pi)
To prove that, you can consider the square of the integral from (-∞,+∞), write that as a double integral and then switch to polar coordinates:
integral over (-∞,+∞) x (-∞,+∞) of
e^[-(x^2 + y^2)] dx dy =
integral over (0,2 pi) x (0,+∞) of
e^[-r^2] r dtheta dr =
2 pi integral over (0,+∞) of
e^[-r^2] r dr = pi
This means that:
integral over (-∞,+∞) of e^(-y^2) dy =
sqrt(pi)
And the integral from zero to infinity is then half this ammount as the integrand is an even function.