**Problem #1**: Show that, for all continuous, integrable functions $ f $ on the interval $ [a,b] $ (plus whatever other conditions you need), we have

$ \displaystyle \left(\int_a^b f(x)dx \right)^2 = \int_a^b \int_a^b f(x) \cdot f(y) dx dy $.

**Problem #2**: Show that, for all continuous, integrable functions $ f $ on the interval $ [0, \infty] $ (plus whatever other conditions you need), we have

$ \displaystyle \int_0^{\infty} \int_0^{\infty} f(x) \cdot f(y) dx dy = \int_0^{\frac{\pi}{2}} \int_0^{\infty} r \cdot g(r, \theta) dr d\theta $

where $ (r, \theta) $ are the polar coordinates for $ (x,y) $ and $ g(r, \theta) = f(x) \cdot f(y) $. Indeed, we notice that $ g(r, \theta) = f(r \cos{\theta}) \cdot f(r \sin{\theta}) $.

**Problem #3**: Use the above two properties to evaluate

$ \displaystyle \int_0^{\infty} e^{-x^2} dx $.

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Comment: Probably one of the coolest tricks in my limited calculus background. The first identity is nice to know and simple to prove, but the second is the really big leap that is useful for evaluating definite integrals which you cannot simply integrate. Hopefully, doing each of these problems, the method will become clear.

The first trick is pretty obvious. Let F = int(b to a) f(x) dx. Then the inside integral is just F * f(y), and the integral of that (with respect to y) is just F * F.

ReplyDeleteThe second trick represents the volume under a surface z = f(x) f(y) constrained on the positive xy-plane. If the positive xy-plane is viewed in polar coordinates, we want the (outer) integrand to represent the volume of a "radial slice" at a given angle theta. (More explicitly: we want to calculate the volume from theta to theta + dtheta.)

This "radial slice" is itself an integral, and cutting the "radial slice" perpendicularly produces a rectangle of width L = r theta = dr theta and height g(r, theta), giving our inside integrand. Integrating this expression gives an integrand for the volume of a "radial slice," which we integrate with respect to theta across the positive xy-plane (that is, from 0 to pi/2).

Oh, I forgot to perform the FINAL TRICK. Let f(x) = e^{-x^2}, right? Okay. So

ReplyDelete(the awesome integral)^2 = (the double integral) = (the radial integral) = int(0 to pi/2) int(0 to infty) r e^{-x^2 - y^2} dr dtheta = int(0 to pi/2) int(0 to infty) r e^{-r^2} dr dtheta = int(0 to pi/2) ( -1/2 e^{-r^2}, r from 0 to infty) dtheta = int(0 to pi/2) (1/2) dtheta = pi/4

so

(the awesome integral) = sqrt(pi) / 2