{"id":491,"date":"2021-04-01T22:20:48","date_gmt":"2021-04-01T22:20:48","guid":{"rendered":"https:\/\/fooledbyrandomnessdotcom.wordpress.com\/?p=491"},"modified":"2021-04-01T22:20:48","modified_gmt":"2021-04-01T22:20:48","slug":"a-peculiar-integral","status":"publish","type":"post","link":"https:\/\/fooledbyrandomness.com\/blog\/index.php\/2021\/04\/01\/a-peculiar-integral\/","title":{"rendered":"A Peculiar Integral"},"content":{"rendered":"\n

Prove<\/p>\n\n\n\n

\\(I= \\displaystyle\\int_{-\\infty }^{\\infty}\\sum_{n=0}^{\\infty } \\frac{\\left(-x^2\\right)^n }{n!^{2 s}}\\; \\mathrm{d}x= \\pi^{1-s}\\).<\/p>\n\n\n\n

We can start as follows, by transforming it into a generalized hypergeometric function:<\/p>\n\n\n\n

\\(I=\\displaystyle\\int_{-\\infty }^{\\infty }\\, _0F_{2 s-1} (\\overbrace{1,1,1,…,1}^{2 s-1 \\text{times}}; -x^2)\\mathrm{d}x\\), since, from the series expansion of the generalized hypergeometric function, \\(\\, _pF_q\\left(a_1,a_p;b_1,b_q;z\\right)=\\sum_{k=0}^{\\infty } \\frac{\\prod_{j=1}^p \\left(a_j\\right)_k z^k}{\\prod_{j=1}^q k! \\left(b_j\\right)_k}\\), where \\((.)_k\\) is the Pochhammer symbol \\((a)_k=\\frac{\\Gamma (a+k)}{\\Gamma (a)}\\).<\/p>\n\n\n\n

<\/p>\n\n\n\n

Now the integrand function does not appear to be convergent numerically, except for \\(s= \\frac{1}{2}\\) where it becomes the Gaussian integral, and the case of \\(s=1\\) where it becomes a Bessel function. For \\(s=\\frac{3}{2}\\) and \\( x=10^{19}\\), the integrand takes values of \\(10^{1015852872356}\\) (serious). Beyond that the computer starts to produce smoke. Yet it eventually converges as there is a closed form solution. It is like saying that it works in theory but not in practice!<\/p>\n\n\n\n

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<\/p>\n<\/div><\/div>\n\n\n\n

<\/p>\n<\/div>\n<\/div>\n\n\n\n

For, it turns out, under the restriction that \\(2 s\\in \\mathbb{Z}_{>\\, 0}\\), we can use the following result:<\/p>\n\n\n\n

\\(\\int_0^{\\infty } t^{\\alpha -1} _pF_q \\left(a_1,\\ldots ,a_p;b_1,\\ldots ,b_q;-t\\right) \\, dt=\\frac{\\Gamma (\\alpha ) \\prod {k=1}^p \\Gamma \\left(a_k-\\alpha \\right)}{\\left(\\prod {k=1}^p \\Gamma \\left(a_k\\right)\\right) \\prod {k=1}^q \\Gamma \\left(b_k-\\alpha \\right)}\\)<\/p>\n\n\n\n

Allora, we can substitute \\(x=\\sqrt(u)\\), and with \\(\\alpha =\\frac{1}{2},p=0,b_k=1,q=2 s-1\\), given that \\(\\Gamma(\\frac{1}{2})=\\sqrt(\\pi)\\), <\/p>\n\n\n\n

\\(I=\\frac{\\sqrt{\\pi }}{\\prod _{k=1}^{2 s-1} \\sqrt{\\pi }}=\\pi ^{1-s}\\).<\/p>\n\n\n\n

So either the integrand eventually converges, or I am doing something wrong, or both. Perhaps neither.<\/p>\n\n\n\n

<\/p>\n\n\n\n

<\/p>\n\n\n\n

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<\/pre>\n\n\n\n
<\/p>\n","protected":false},"excerpt":{"rendered":"
Prove \\(I= \\displaystyle\\int_{-\\infty }^{\\infty}\\sum_{n=0}^{\\infty } \\frac{\\left(-x^2\\right)^n }{n!^{2 s}}\\; \\mathrm{d}x= \\pi^{1-s}\\). We can start as follows, by transforming it into a generalized hypergeometric function: \\(I=\\displaystyle\\int_{-\\infty }^{\\infty }\\, _0F_{2 s-1} (\\overbrace{1,1,1,…,1}^{2 s-1 \\text{times}}; -x^2)\\mathrm{d}x\\), since, from the series expansion of the generalized hypergeometric function, \\(\\, _pF_q\\left(a_1,a_p;b_1,b_q;z\\right)=\\sum_{k=0}^{\\infty } \\frac{\\prod_{j=1}^p \\left(a_j\\right)_k z^k}{\\prod_{j=1}^q k! \\left(b_j\\right)_k}\\), where \\((.)_k\\) is the Pochhammer … Continue reading “A Peculiar Integral”<\/span><\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_bbp_topic_count":0,"_bbp_reply_count":0,"_bbp_total_topic_count":0,"_bbp_total_reply_count":0,"_bbp_voice_count":0,"_bbp_anonymous_reply_count":0,"_bbp_topic_count_hidden":0,"_bbp_reply_count_hidden":0,"_bbp_forum_subforum_count":0,"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[1],"tags":[],"class_list":["post-491","post","type-post","status-publish","format-standard","hentry","category-uncategorized"],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/fooledbyrandomness.com\/blog\/index.php\/wp-json\/wp\/v2\/posts\/491","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/fooledbyrandomness.com\/blog\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/fooledbyrandomness.com\/blog\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/fooledbyrandomness.com\/blog\/index.php\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/fooledbyrandomness.com\/blog\/index.php\/wp-json\/wp\/v2\/comments?post=491"}],"version-history":[{"count":0,"href":"https:\/\/fooledbyrandomness.com\/blog\/index.php\/wp-json\/wp\/v2\/posts\/491\/revisions"}],"wp:attachment":[{"href":"https:\/\/fooledbyrandomness.com\/blog\/index.php\/wp-json\/wp\/v2\/media?parent=491"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/fooledbyrandomness.com\/blog\/index.php\/wp-json\/wp\/v2\/categories?post=491"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/fooledbyrandomness.com\/blog\/index.php\/wp-json\/wp\/v2\/tags?post=491"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}