{"id":108,"date":"2016-05-08T01:53:19","date_gmt":"2016-05-08T01:53:19","guid":{"rendered":"http:\/\/www.laurentlessard.com\/bookproofs\/?p=108"},"modified":"2019-12-22T12:12:15","modified_gmt":"2019-12-22T18:12:15","slug":"a-clever-integral","status":"publish","type":"post","link":"https:\/\/laurentlessard.com\/bookproofs\/a-clever-integral\/","title":{"rendered":"A clever integral"},"content":{"rendered":"

<\/p>I was recently reminded of this problem from one of my favorite books: Problem-Solving Through Problems<\/a>. The problem originally appeared in the 1980 Putnam Competition<\/a>.\n

Evaluate the following definite integral.<\/p>\n

\\[
\n\\int_0^{\\pi\/2} \\frac{\\mathrm{d}x}{1 + (\\tan x)^{\\sqrt{2}}}
\n\\]<\/p>\n

The solution:
\n [Show Solution]<\/a><\/p>\n

The integrand doesn\u2019t have an obvious antiderivative and it\u2019s not clear how one would go about computing it. So what can we do? In this case, the limits of integration play a key role. Let\u2019s call our problematic integral $J_1$. Note that:<\/p>\n

\\[
\nJ_1 = \\int_0^{\\pi\/2} \\frac{\\mathrm{d}x}{1 + (\\tan x)^{\\sqrt{2}}} = \\int_0^{\\pi\/2} \\frac{(\\cos x)^{\\sqrt{2}}}{(\\cos x)^{\\sqrt{2}} + (\\sin x)^{\\sqrt{2}}} \\,\\mathrm{d}x
\n\\]<\/p>\n

No progress yet, but the new form of $J_1$ suggests a certain symmetry. Specifically, define:<\/p>\n

\\[
\nJ_2 = \\int_0^{\\pi\/2} \\frac{(\\sin x)^{\\sqrt{2}}}{(\\cos x)^{\\sqrt{2}} + (\\sin x)^{\\sqrt{2}}} \\,\\mathrm{d}x
\n\\]<\/p>\n

Two key observations. First, $J_1 + J_2 = \\pi\/2$. This is clear because the integrands of $J_1$ and $J_2$ sum to 1. Second, $J_1=J_2$. This is clear because if we make the substitution $x \\mapsto \\frac{\\pi}{2} \u2013 x$, then $\\cos$ and $\\sin$ trade places and the integral remains unchanged. These facts together imply that<\/p>\n

\\[
\nJ_1 = J_2 = \\frac{\\pi}{4}
\n\\]<\/p>\n

Here is a plot showing the integral represented as an area under the curve. This area is precisely half the area of the entire rectangle.<\/p>\n

$\"integrand$ <\/p>\n

Note also that the $\\sqrt{2}$ was a red herring<\/a>. We could have replaced it by some other real number and the value of the integral would be unchanged!<\/p>\n

We managed to evaluate the integral without ever computing the antiderivative. This isn\u2019t uncommon \u2014 for example, a standard integral that shows up when working with normal distributions is:<\/p>\n

\\[
\n\\int_{-\\infty}^\\infty e^{-x^2}\\,\\mathrm{d}x = \\sqrt{\\pi}
\n\\]<\/p>\n

This integral can be evaluated even though $e^{-x^2}$ doesn\u2019t have an antiderivative\u2026 but that\u2019s a topic for another post!<\/p>\n<\/div>\n

<\/p>\n<\/body>","protected":false},"excerpt":{"rendered":"

I was recently reminded of this problem from one of my favorite books: Problem-Solving Through Problems. The problem originally appeared in the 1980 Putnam Competition. Evaluate the following definite integral. \\[ \\int_0^{\\pi\/2} \\frac{\\mathrm{d}x}{1 + (\\tan x)^{\\sqrt{2}}} \\] The solution: [Show Solution] The integrand doesn\u2019t have an obvious antiderivative and it\u2019s not clear how one would … Continue reading “A clever integral”<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"_uf_show_specific_survey":0,"_uf_disable_surveys":false,"footnotes":""},"categories":[13],"tags":[4,14],"class_list":["post-108","post","type-post","status-publish","format-standard","hentry","category-competition","tag-integration","tag-symmetry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/posts\/108","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/comments?post=108"}],"version-history":[{"count":21,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/posts\/108\/revisions"}],"predecessor-version":[{"id":2731,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/posts\/108\/revisions\/2731"}],"wp:attachment":[{"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/media?parent=108"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/categories?post=108"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/tags?post=108"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}