The following integral may be problematic for a fresh calculus student, even if armed with a list of antiderivatives:

INTEGRAL_{0 to infinity} exp(-x²) dx.

Why? Well, there isn't a closed-form expression for the antiderivative of the integrand, so the Fundamental Theorem of Calculus won't help. But the expression is meaningful, since the it represents the area under the curve from 0 to infinity.

Furthermore, there is a nice trick to find the answer! Call the integral I. Multiply the integral by itself: this gives

I² = [ INTEGRAL_{0 to infinity} exp(-x²) dx ] [ INTEGRAL_{0 to infinity} exp(-y²) dy ]

then view as an integral over the first quadrant in the plane:

= [ INTEGRAL_{0 to infinity} INTEGRAL_{0 to infinity} exp(-x²-y²) dx dy]

then change to polar coordinates(!):

= INTEGRAL_{0 to Pi/2} INTEGRAL_{0 to infinity} exp(-r²) r dr d(THETA).

Now this is quite easy to evaluate: you find that I²=Pi/4.

This means that I, the original value of the

integral you were looking for, is Sqrt[Pi]/2.

Wow!

This trick is often learned in multivariable calculus course; it is best to show it right after learning to integrate in polar coordinates

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