Integrals Involving a Parameter

Abstract

Consider the integral¹

EquationSource% MathType!MTEF!2!1!+- % feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOqamaaBa % aaleaacaWGtbaabeaakiaacIcacaWG4bGaaiykaiabg2da9maalaaa % baGaaGymaaqaaiaadohacaGGHaaaamaapedabaWaaSaaaeaacaWG1b % WaaWbaaSqabeaacaWGZbaaaOGaamizaiaadwhaaeaacaWGLbWaaWba % aSqabeaacaWG1bGaey4kaSIaamiwaaaakiabgkHiTiaaigdaaaGaai % ilaaWcbaGaaGimaaqaaiabg6HiLcqdcqGHRiI8aaaa!4C2F!]]</EquationSource><EquationSource Format="TEX"><![CDATA[$${B_S}(x) = \frac{1}{{s!}}\int_0^\infty {\frac{{{u^s}du}}{{{e^{u + X}} - 1}},} $$

(1)

and suppose that we require an expansion for small values of the parameter x. When x = 0, the integral is simply a zeta function. If we attempt to find an expansion for small x by expanding the integrand in powers of x directly, the expansion will ultimately break down. To see this explicitly, suppose for simplicity that 0 < s < 1. We have then

EquationSource% MathType!MTEF!2!1!+- % feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOqamaaBa % aaleaacaWGtbaabeaakiaacIcacaWG4bGaaiykaiabg2da9maalaaa % baGaaGymaaqaaiaadohacaGGHaaaaiaacUhadaWdXaqaamaalaaaba % GaamyDamaaCaaaleqabaGaam4CaaaakiaadsgacaWG1baabaGaamyz % amaaCaaaleqabaGaamyDaaaakiabgkHiTiaaigdaaaGaey4kaSYaa8 % qmaeaacaWG1bWaaWbaaSqabeaacaWGZbaaaOGaamizaiaadwhaaSqa % aiaaicdaaeaacqGHEisPa0Gaey4kIipaaSqaaiaaicdaaeaacqGHEi % sPa0Gaey4kIipakiaacUfadaWcaaqaaiaaigdaaeaacaWGLbWaaWba % aSqabeaacaWG1bGaey4kaSIaamiwaaaakiabgkHiTiaaigdaaaGaey % OeI0YaaSaaaeaacaaIXaaabaGaamyzamaaCaaaleqabaGaamyDaaaa % kiabgkHiTiaaigdaaaGaaiyxaiaac2hacqGH9aqpcqaHcpGvcaGGOa % Gaam4uaiabgUcaRiaaigdacaGGPaGaeyOeI0IaamiEamaapedabaWa % aSaaaeaacaWG1bWaaWbaaSqabeaacaWGZbaaaOGaamizaiaadwhaae % aacaGGOaGaamyzamaaCaaaleqabaGaamyDaaaakiabgkHiTiaaigda % caGGPaWaaWbaaSqabeaacaaIYaaaaaaaaeaacaaIWaaabaGaeyOhIu % kaniabgUIiYdGccqGHRaWkcaWGpbGaaiikaiaadIhadaahaaWcbeqa % aiaaikdaaaGccaGGPaGaaiOlaaaa!7F1C!]]</EquationSource><EquationSource Format="TEX"><![CDATA[$${B_S}(x) = \frac{1}{{s!}}\{ \int_0^\infty {\frac{{{u^s}du}}{{{e^u} - 1}} + \int_0^\infty {{u^s}du} } [\frac{1}{{{e^{u + X}} - 1}} - \frac{1}{{{e^u} - 1}}]\} = \varsigma (S + 1) - x\int_0^\infty {\frac{{{u^s}du}}{{{{({e^u} - 1)}^2}}}} + O({x^2}).$$

(2)