Question

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?

Answer 1

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.