The following integral may be problematic for a freshman calculus student, even if armed with a list of antiderivatives:<\/p>\n\n\n\n
INTEGRAL0 to infinity<\/sub> exp(-x2<\/sup>) dx.<\/p>\n\n\n\n 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.<\/p>\n\n\n\n Furthermore, there is a nice trick to find the answer! Call the integral I. Multiply the integral by itself: this gives<\/p>\n\n\n\n I2<\/sup> = [ INTEGRAL0 to infinity<\/sub> exp(-x2<\/sup>) dx ] [ INTEGRAL0 to infinity<\/sub> exp(-y2<\/sup>) dy ]<\/p>\n\n\n\n then view as an integral over the first quadrant in the plane:<\/p>\n\n\n\n = [ INTEGRAL0 to infinity<\/sub> INTEGRAL0 to infinity<\/sub> exp(-x2<\/sup>-y2<\/sup>) dx dy]<\/p>\n\n\n\n then change to polar coordinates (!):<\/p>\n\n\n\n = INTEGRAL0 to Pi\/2<\/sub> INTEGRAL0 to infinity<\/sub> exp(-r2<\/sup>) r dr d(THETA).<\/p>\n\n\n\n Now this is quite easy to evaluate: you find that I2<\/sup>=Pi\/4. This means that I, the original value of the integral you were looking for, is Sqrt[Pi]\/2.<\/p>\n\n\n\n Wow!<\/p>\n\n\n\n Presentation Suggestions:<\/strong> The Math Behind the Fact:<\/strong> How to Cite this Page:<\/strong>\u00a0 Fun Fact suggested by:<\/strong> The following integral may be problematic for a freshman calculus student, even if armed with a list of antiderivatives: INTEGRAL0...<\/p>\n","protected":false},"author":7,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"tags":[20,7,3,4,123],"class_list":["post-471","page","type-page","status-publish","hentry","tag-analysis","tag-calculus","tag-easy","tag-medium","tag-multivariable-calculus"],"jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/pages\/471","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/users\/7"}],"replies":[{"embeddable":true,"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/comments?post=471"}],"version-history":[{"count":3,"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/pages\/471\/revisions"}],"predecessor-version":[{"id":1486,"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/pages\/471\/revisions\/1486"}],"wp:attachment":[{"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/media?parent=471"}],"wp:term":[{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/tags?post=471"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
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.<\/p>\n\n\n\n
You may recognize the integrand as the familiar (unscaled) bell curve<\/em>. 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<\/em>.<\/p>\n\n\n\n
Su, Francis E., et al. “Impossible Integral?.”\u00a0Math Fun Facts<\/em>. <http:\/\/www.math.hmc.edu\/funfacts>.<\/p>\n\n\n\n
Lesley Ward<\/p>\n","protected":false},"excerpt":{"rendered":"