Impossible Integral?

The following integral may be problematic for a freshman 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!

Presentation Suggestions:
This trick is often learned in multivariable calculus course; it is best to show it right after learning to integrate in polar coordinates. If polar coordinates have not been introduced yet, you can view the squared integral as the volume of a solid of revolution, and evaluate using shells.

The Math Behind the Fact:
You may recognize the integrand as the familiar (unscaled) bell curve. An alternate way of evaluating this integral (without appealing to an unmotivated trick!) is to view it as a complex integral and use residue theory. You can learn more about this in a course on complex analysis.

How to Cite this Page: 
Su, Francis E., et al. “Impossible Integral?.” Math Fun Facts. <https://www.math.hmc.edu/funfacts>.

Fun Fact suggested by:
Lesley Ward