The following integral may be problematic for
a freshman calculus student, even if armed with
a list of antiderivatives:
INTEGRAL_{0 to infinity} exp(x^{2}) dx.
Why? Well, there isn't a closedform 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
I^{2} =
[ INTEGRAL_{0 to infinity} exp(x^{2}) dx ]
[ INTEGRAL_{0 to infinity} exp(y^{2}) dy ]
then view as an integral over the first quadrant in
the plane:
= [ INTEGRAL_{0 to infinity}
INTEGRAL_{0 to infinity}
exp(x^{2}y^{2}) dx dy]
then change to polar coordinates
(!):
= INTEGRAL_{0 to Pi/2}
INTEGRAL_{0 to infinity}
exp(r^{2}) r dr d(THETA).
Now this is quite easy to evaluate: you find that
I^{2}=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.
<http://www.math.hmc.edu/funfacts>.

