Convergent Integrals

SPECIES

exp(2πt) – 1 as the denominator

$\displaystyle \int_0^{\infty_+} \dfrac{1}{e^{2\pi t}-1} \, dt = \dfrac{\zeta(1)}{2\pi} = \dfrac{\gamma}{2\pi} \quad \text{(divergent)}$

$\displaystyle \int_0^{\infty_+} \dfrac{t}{e^{2\pi t}-1} \, dt = \dfrac{\zeta(2)}{4\pi^2} = \dfrac{1}{24}$

$\displaystyle \int_0^{\infty_+} \dfrac{t^2}{e^{2\pi t}-1} \, dt= \dfrac{\zeta(3)}{4\pi^3}$

$\displaystyle \int_0^{\infty_+} \dfrac{t^3}{e^{2\pi t}-1} \, dt = \dfrac{3\,\zeta(4)}{8\pi^4} = \dfrac{1}{240}$

$\displaystyle \int_0^{\infty_+} \dfrac{t^4}{e^{2\pi t}-1} \, dt = \dfrac{3\,\zeta(5)}{4\pi^5}$

$\displaystyle \int_0^{\infty_+} \dfrac{t^5}{e^{2\pi t}-1} \, dt = \dfrac{15\,\zeta(6)}{8\pi^6} = \dfrac{1}{504}$

$\displaystyle \int_0^{\infty_+} \dfrac{t^6}{e^{2\pi t}-1} \, dt = \dfrac{45\,\zeta(7)}{8\pi^7}$

$\displaystyle \int_0^{\infty_+} \dfrac{t^7}{e^{2\pi t}-1} \, dt = \dfrac{315\,\zeta(8)}{16\pi^8} = \dfrac{1}{480}$

$\displaystyle \int_0^{\infty_+} \dfrac{t^8}{e^{2\pi t}-1} \, dt = \dfrac{315\,\zeta(9)}{4\pi^9}$

$\displaystyle \int_0^{\infty_+} \dfrac{t^9}{e^{2\pi t}-1} \, dt = \dfrac{2835\,\zeta(10)}{8\pi^{10}} = \dfrac{1}{264}$

$\displaystyle \int_0^{\infty_+} \dfrac{t^{10}}{e^{2\pi t}-1} \, dt = \dfrac{14175\,\zeta(11)}{8\pi^{11}}$

$\displaystyle \int_0^{\infty_+} \dfrac{\sin t}{e^{2\pi t}-1} \, dt = \dfrac{1}{4} \left(\coth\dfrac{1}{2} - 2\right) = \dfrac{1}{4}\coth\dfrac{1}{2} - \dfrac{1}{2}$

$\displaystyle \int_0^{\infty_+} \dfrac{\sinh t}{e^{2\pi t}-1} \, dt = \dfrac{1}{4} \left(2 - \cot\dfrac{1}{2}\right) = \dfrac{1}{2} - \dfrac{1}{4}\cot\dfrac{1}{2}$

$i\,\displaystyle \int_0^{\infty_+} \dfrac{\ln(1\!+\!it) - \ln(1\!-\!it)}{e^{2\pi t}-1} \, dt = \dfrac{\ln\pi}{2} + \dfrac{\ln2}{2} - 1$

$i\,\displaystyle \int_0^{\infty_+} \dfrac{\ln(2\!+\!it) - \ln(2\!-\!it)}{e^{2\pi t}-1} \, dt = \dfrac{\ln\pi}{2} + 2\ln2 - 2$

$i\,\displaystyle \int_0^{\infty_+} \dfrac{\ln(3\!+\!it) - \ln(3\!-\!it)}{e^{2\pi t}-1} \, dt = \dfrac{\ln\pi}{2} - \dfrac{\ln2}{2} + \dfrac{5\ln3}{2} - 3$

$i\,\displaystyle \int_0^{\infty_+} \dfrac{\ln(4\!+\!it) - \ln(4\!-\!it)}{e^{2\pi t}-1} \, dt = \dfrac{\ln\pi}{2} - \dfrac{13\ln2}{2} - \ln3 - 4$

$i\,\displaystyle \int_0^{\infty_+} \dfrac{\ln(5\!+\!it) - \ln(5\!-\!it)}{e^{2\pi t}-1} \, dt = \dfrac{\ln\pi}{2} - \dfrac{5\ln2}{2} - \ln3 + \dfrac{9\ln5}{2} - 5$

$i\,\displaystyle \int_0^{\infty_+} \dfrac{\ln(6\!+\!it) - \ln(6\!-\!it)}{e^{2\pi t}-1} \, dt = \dfrac{\ln\pi}{2} + 3\ln2 + \dfrac{9\ln3}{2} - \ln5 - 6$

$i\,\displaystyle \int_0^{\infty_+} \dfrac{\dfrac{\ln(1\!+\!it)}{1\!+\!it} - \dfrac{\ln(1\!-\!it)}{1\!-\!it}}{e^{2\pi t}-1} \, dt = \gamma_1$

$i\,\displaystyle \int_0^{\infty_+} \dfrac{\dfrac{\ln(2\!+\!it)}{2\!+\!it} - \dfrac{\ln(2\!-\!it)}{2\!-\!it}}{e^{2\pi t}-1} \, dt = \gamma_1 - \dfrac{\ln 2}{4} + \dfrac{\ln^2 2}{2}$

$i\,\displaystyle \int_0^{\infty_+} \dfrac{\dfrac{\ln(3\!+\!it)}{3\!+\!it} - \dfrac{\ln(3\!-\!it)}{3\!-\!it}}{e^{2\pi t}-1} \, dt = \gamma_1 - \dfrac{\ln 2}{2} - \dfrac{\ln 3}{6} + \dfrac{\ln^2 3}{2}$

$i\,\displaystyle \int_0^{\infty_+} \dfrac{\ln^2(1\!+\!it) - \ln^2(1\!-\!it)}{e^{2\pi t}-1} \, dt \\ = 2 + \zeta''(0) = 2 - \dfrac{\ln^2(2\pi)}{2} - \dfrac{\pi^2}{24} + \dfrac{\gamma^2}{2} + \gamma_1$

$i\,\displaystyle \int_0^{\infty_+} \dfrac{\ln^2(2\!+\!it) - \ln^2(2\!-\!it)}{e^{2\pi t}-1} \, dt \\ = \zeta''(0) - \dfrac{\ln^2(2)}{2} + 2(2 - 2\ln 2 + \ln^2(2)) \\ = 4 - \dfrac{\pi^2}{24} - \ln 2 \left(4 - \dfrac{3\ln 2}{2}\right) - \dfrac{\ln^2(2\pi)}{2} + \dfrac{\gamma^2}{2} + \gamma_1$

exp(2πt) – 1 in the denominator

$\displaystyle \int_0^{\infty_+} \dfrac{t}{(1+t^2) \, (e^{2\pi t}\!-\!1)} \, dt = \dfrac{1}{4} + \dfrac{\zeta(1)}{2} = \dfrac{1}{4} + \dfrac{\gamma}{2}$

$\displaystyle \int_0^{\infty_+} \dfrac{t}{(1+t^2)^2 \, (e^{2\pi t}\!-\!1)} \, dt = -\dfrac{3}{8} + \dfrac{\zeta(2)}{4} = -\dfrac{3}{8} + \dfrac{\pi^2}{24}$

$\displaystyle \int_0^{\infty_+} \dfrac{t}{(1+t^2)^3 \, (e^{2\pi t}\!-\!1)} \, dt = -\dfrac{7}{32} + \dfrac{\zeta(2)}{16} + \dfrac{\zeta(3)}{8} \\ = -\dfrac{7}{32} + \dfrac{\pi^2}{96} + \dfrac{\zeta(3)}{8}$

$\displaystyle \int_0^{\infty_+} \dfrac{t^3}{(1+t^2)^3 \, (e^{2\pi t}\!-\!1)} \, dt = -\dfrac{5}{32} + \dfrac{3\zeta(2)}{16} - \dfrac{\zeta(3)}{8} \\ = -\dfrac{5}{32} + \dfrac{\pi^2}{32} - \dfrac{\zeta(3)}{8}$

$\displaystyle \int_0^{\infty_+} \dfrac{t(t^2\!-\!3)}{(1+t^2)^3 \, (e^{2\pi t}\!-\!1)} \, dt = \dfrac{1}{2} - \dfrac{\zeta(3)}{2}$

exp(2πt) + 1 as the denominator

$\displaystyle \int_0^{\infty_+} \dfrac{1}{e^{2\pi t}+1} \, dt = \dfrac{(1-2^0)\,\zeta(1)}{2\pi} = \dfrac{\eta(1)}{2\pi} = \dfrac{\ln2}{2\pi} \quad \text{(divergent)}$

$\displaystyle \int_0^{\infty_+} \dfrac{t}{e^{2\pi t}+1} \, dt = \dfrac{\zeta(2)}{8\,\pi^2} = \dfrac{\eta(2)}{4\,\pi^2} = \dfrac{1}{48}$

$\displaystyle \int_0^{\infty_+} \dfrac{t^2}{e^{2\pi t}+1} \, dt = \dfrac{3\,\zeta(3)}{16\,\pi^3} = \dfrac{\eta(3)}{4\,\pi^3}$

$\displaystyle \int_0^{\infty_+} \dfrac{t^3}{e^{2\pi t}+1} \, dt = \dfrac{21\,\zeta(4)}{64\,\pi^4} = \dfrac{3\,\eta(4)}{8\,\pi^4} = \dfrac{7}{1920}$

$\displaystyle \int_0^{\infty_+} \dfrac{t^4}{e^{2\pi t}+1} \, dt = \dfrac{45\,\zeta(5)}{64\,\pi^5} = \dfrac{3\,\eta(5)}{4\,\pi^5}$

$\displaystyle \int_0^{\infty_+} \dfrac{t^5}{e^{2\pi t}+1} \, dt = \dfrac{465\,\zeta(6)}{256\,\pi^6} = \dfrac{15\,\eta(6)}{8\,\pi^6} = \dfrac{31}{16128}$

$\displaystyle \int_0^{\infty_+} \dfrac{t^6}{e^{2\pi t}+1} \, dt = \dfrac{2835\,\zeta(7)}{512\,\pi^7} = \dfrac{45\,\eta(7)}{8\,\pi^7}$

$\displaystyle \int_0^{\infty_+} \dfrac{t^7}{e^{2\pi t}+1} \, dt = \dfrac{40005\,\zeta(8)}{2048\,\pi^8} = \dfrac{315\,\eta(8)}{16\,\pi^8} = \dfrac{127}{61440}$

$\displaystyle \int_0^{\infty_+} \dfrac{t^8}{e^{2\pi t}+1} \, dt = \dfrac{80325\,\zeta(9)}{1024\,\pi^9} = \dfrac{315\,\eta(9)}{4\,\pi^9}$

$\displaystyle \int_0^{\infty_+} \dfrac{t^9}{e^{2\pi t}+1} \, dt = \dfrac{1448685\,\zeta(10)}{4096\,\pi^{10}} = \dfrac{2835\,\eta(10)}{8\,\pi^{10}} = \dfrac{511}{135168}$

$\displaystyle \int_0^{\infty_+} \dfrac{t^{10}}{e^{2\pi t}+1} \, dt = \dfrac{14501025\,\zeta(11)}{8192\,\pi^{11}} = \dfrac{14175\,\eta(11)}{8\,\pi^{11}}$

sinh(πt) as the denominator

$\displaystyle i\int_{\infty_-}^{\infty_+} \dfrac{e^{it}-e^{-it}}{\sinh(\pi t)} \, dt = -2\tanh\!\left(\dfrac{1}{2}\right)$

$\displaystyle i\int_{\infty_-}^{\infty_+} \dfrac{\ln(1+it)}{\sinh(\pi t)} \, dt = \ln\!\left(\dfrac{2}{\pi}\right) = -\ln\pi + \ln2$

$\displaystyle i\int_{\infty_-}^{\infty_+} \dfrac{\ln(1-it)}{\sinh(\pi t)} \, dt = \ln\!\left(\dfrac{\pi}{2}\right) = \ln\pi - \ln2$

$\displaystyle i\int_{\infty_-}^{\infty_+} \dfrac{\ln(1+it) - \ln(1-it)}{\sinh(\pi t)} \, dt = 2\ln\!\left(\dfrac{2}{\pi}\right) = -2\ln\pi + 2\ln2$

$\displaystyle i\int_{\infty_-}^{\infty_+} \dfrac{\ln(2+it) - \ln(2-it)}{\sinh(\pi t)} \, dt = 2\ln\!\left(\dfrac{\pi}{4}\right) = 2\ln\pi - 4\ln2$

$\displaystyle i\int_{\infty_-}^{\infty_+} \dfrac{\ln(3+it) - \ln(3-it)}{\sinh(\pi t)} \, dt = 2\ln\!\left(\dfrac{8}{3\pi}\right) = - 2\ln\pi + 6\ln2 - \nolinebreak 2\ln3$

$\displaystyle i\int_{\infty_-}^{\infty_+} \dfrac{\ln(4+it) - \ln(4-it)}{\sinh(\pi t)} \, dt = 2\ln\!\left(\dfrac{9\pi}{32}\right) = 2\ln\pi - 10\ln2 + \nolinebreak 4\ln3$

$\displaystyle i\int_{\infty_-}^{\infty_+} \dfrac{\ln(5+it) - \ln(5-it)}{\sinh(\pi t)} \, dt \\ = 2\ln\!\left(\dfrac{128}{45\pi}\right) = -2\ln\pi + 14\ln2 - 4\ln3 - 2\ln5$

cosh(πt) as the denominator

$\displaystyle \int_{\infty_-}^{\infty_+} \dfrac{e^{it}}{\cosh(\pi t)} \, dt = \int_{\infty_-}^{\infty_+} \dfrac{e^{-it}}{\cosh(\pi t)} \, dt = \dfrac{2\sqrt{e}}{e+1}$

$\displaystyle \int_{\infty_-}^{\infty_+} \dfrac{e^{it}-e^{-it}}{\cosh(\pi t)} \, dt = 0$

$\displaystyle \int_{\infty_-}^{\infty_+} \dfrac{\ln(\frac{1}{2}+it)}{\cosh(\pi t)} \, dt = \int_{\infty_-}^{\infty_+} \dfrac{\ln(\frac{1}{2}-it)}{\cosh(\pi t)} \, dt = \ln\!\left(\dfrac{2}{\pi}\right) = -\ln\pi + \ln2$

$\displaystyle \int_{\infty_-}^{\infty_+} \dfrac{\ln(\frac{1}{2}+it) - \ln(\frac{1}{2}-it)}{\cosh(\pi t)} \, dt = 0$

$\displaystyle \int_{\infty_-}^{\infty_+} \dfrac{\ln(\frac{3}{2}+it)}{\cosh(\pi t)} \, dt = \int_{\infty_-}^{\infty_+} \dfrac{\ln(\frac{3}{2}-it)}{\cosh(\pi t)} \, dt = \ln\!\left(\dfrac{\pi}{2}\right) = \ln\pi - \ln2$

$\displaystyle \int_{\infty_-}^{\infty_+} \dfrac{\ln(\frac{3}{2}+it) - \ln(\frac{3}{2}-it)}{\cosh(\pi t)} \, dt = 0$

$\displaystyle \int_{\infty_-}^{\infty_+} \dfrac{\ln(\frac{5}{2}+it)}{\cosh(\pi t)} \, dt = \int_{\infty_-}^{\infty_+} \dfrac{\ln(\frac{5}{2}-it)}{\cosh(\pi t)} \, dt = \ln\!\left(\dfrac{8}{\pi}\right) = -\ln\pi + \nolinebreak 3\ln2$

$\displaystyle \int_{\infty_-}^{\infty_+} \dfrac{\ln(\frac{5}{2}+it) - \ln(\frac{5}{2}-it)}{\cosh(\pi t)} \, dt = 0$

$\displaystyle \int_{\infty_-}^{\infty_+} \dfrac{\ln(\frac{7}{2}+it)}{\cosh(\pi t)} \, dt = \int_{\infty_-}^{\infty_+} \dfrac{\ln(\frac{7}{2}-it)}{\cosh(\pi t)} \, dt \\ = \ln\!\left(\dfrac{9\pi}{8}\right) = \ln\pi - 3\ln2 + 2\ln3$

$\displaystyle \int_{\infty_-}^{\infty_+} \dfrac{\ln(\frac{7}{2}+it) - \ln(\frac{7}{2}-it)}{\cosh(\pi t)} \, dt = 0$

$\displaystyle \int_{\infty_-}^{\infty_+} \dfrac{\ln(\frac{9}{2}+it)}{\cosh(\pi t)} \, dt = \int_{\infty_-}^{\infty_+} \dfrac{\ln(\frac{9}{2}-it)}{\cosh(\pi t)} \, dt \\ = \ln\!\left(\dfrac{128}{9\pi}\right) = -\ln\pi + 7\ln2 - 2\ln3$

$\displaystyle \int_{\infty_-}^{\infty_+} \dfrac{\ln(\frac{9}{2}+it) - \ln(\frac{9}{2}-it)}{\cosh(\pi t)} \, dt = 0$

SUBSPECIES

Pages: 1 2 3 4 5