Divergent Integrals

SPECIES

Monomial Integrals

$\displaystyle \int_0^{\infty_+} dt = -1$

$\displaystyle \int_0^{\infty_+} \sqrt{t} \, dt = -\dfrac{2i}{3}$

$\displaystyle \int_0^{\infty_+} t \, dt = \dfrac{1}{2}$

$\displaystyle \int_0^{\infty_+} t^2 \, dt = -\dfrac{1}{3}$

$\displaystyle \int_0^{\infty_+} t^3 \, dt = \dfrac{1}{4}$

$\displaystyle \int_0^{\infty_+} t^4 \, dt = -\dfrac{1}{5}$

Monomial Ratio Integrals

$\displaystyle \int_0^{\infty_+} \dfrac{1}{t} \, dt = \underbrace{\ln \infty_+}_{0} - \underbrace{\ln 0}_{-\gamma} = \gamma$

Polynomial Integrals

$\displaystyle \int_0^{\infty_+} (1+t) \, dt = -\dfrac{1}{2}$

$\displaystyle \int_0^{\infty_+} (1+t+t^2) \, dt = -\dfrac{5}{6}$

$\displaystyle \int_0^{\infty_+} (1+t+t^2+t^3) \, dt = -\dfrac{7}{12}$

Polynomial Ratio Integrals

$\displaystyle \int_0^{\infty_+} \dfrac{1}{1-t} \, dt = -\ln 2$

$\displaystyle \int_0^{\infty_+} \dfrac{1}{1+t} \, dt = \ln 0 = -\gamma$

$\displaystyle \int_0^{\infty_+} \dfrac{t}{1+t} \, dt = -1 - \ln 0 = -1 + \gamma$

$\displaystyle \int_0^{\infty_+} \dfrac{t^2}{1+t} \, dt = \dfrac{3}{2} + \ln 0 = \dfrac{3}{2} - \gamma$

$\displaystyle \int_0^{\infty_+} \dfrac{t^3}{1+t} \, dt = -\dfrac{11}{6} - \ln 0 = -\dfrac{11}{6} + \gamma$

$\displaystyle \int_0^{\infty_+} \dfrac{t^4}{1+t} \, dt = \dfrac{25}{12} + \ln 0 = \dfrac{25}{12} - \gamma$

$\displaystyle \int_0^{\infty_+} \dfrac{t^5}{1+t} \, dt = -\dfrac{137}{60} - \ln 0 = -\dfrac{137}{60} + \gamma$

$\displaystyle \int_0^{\infty_+} \dfrac{t^6}{1+t} \, dt = \dfrac{49}{20} + \ln 0 = \dfrac{49}{20} - \gamma$

$\displaystyle \int_0^1 \dfrac{1}{1-t} \, dt = \int_0^1 \dfrac{1}{t} \, dt = \gamma - H_0 = \gamma$

$\displaystyle \int_0^1 \dfrac{t}{1-t} \, dt = \int_0^1 \dfrac{1-t}{t} \, dt = \gamma - H_1 = \gamma - 1$

$\displaystyle \int_0^1 \dfrac{t^2}{1-t} \, dt = \int_0^1 \dfrac{(1-t)^2}{t} \, dt = \gamma - H_2 = \gamma - \dfrac{3}{2}$

$\displaystyle \int_0^1 \dfrac{t^3}{1-t} \, dt = \int_0^1 \dfrac{(1-t)^3}{t} \, dt = \gamma - H_3 = \gamma - \dfrac{11}{6}$

$\displaystyle \int_0^1 \dfrac{t^4}{1-t} \, dt = \int_0^1 \dfrac{(1-t)^4}{t} \, dt = \gamma - H_4 = \gamma - \dfrac{25}{12}$

$\displaystyle \int_0^1 \dfrac{t^5}{1-t} \, dt = \int_0^1 \dfrac{(1-t)^5}{t} \, dt = \gamma - H_5 = \gamma - \dfrac{137}{60}$

$\displaystyle \int_0^1 \dfrac{t^6}{1-t} \, dt = \int_0^1 \dfrac{(1-t)^6}{t} \, dt = \gamma - H_6 = \gamma - \dfrac{49}{20}$

$\displaystyle \int_0^1 \dfrac{1}{(1-t)^2} \, dt = \int_0^1 \dfrac{1}{t^2} \, dt = -\dfrac{1}{2}$

$\displaystyle \int_0^1 \dfrac{t}{(1-t)^2} \, dt = \int_0^1 \dfrac{1-t}{t^2} \, dt = -\dfrac{1}{2} + H_0^{(2)} - \gamma = -\dfrac{1}{2} - \gamma$

$\displaystyle \int_0^1 \dfrac{t^2}{(1-t)^2} \, dt = \int_0^1 \dfrac{(1-t)^2}{t^2} \, dt = -\dfrac{1}{2} + H_1^{(2)} - 2\gamma = \dfrac{1}{2} - 2\gamma$

$\displaystyle \int_0^1 \dfrac{t^3}{(1-t)^2} \, dt = \int_0^1 \dfrac{(1-t)^3}{t^2} \, dt = -\dfrac{1}{2} + H_2^{(2)} - 3\gamma = 2 - 3\gamma$

$\displaystyle \int_0^1 \dfrac{t^4}{(1-t)^2} \, dt = \int_0^1 \dfrac{(1-t)^4}{t^2} \, dt = -\dfrac{1}{2} + H_3^{(2)} - 4\gamma = \dfrac{23}{6} - 4\gamma$

$\displaystyle \int_0^1 \dfrac{t^5}{(1-t)^2} \, dt = \int_0^1 \dfrac{(1-t)^5}{t^2} \, dt = -\dfrac{1}{2} + H_4^{(2)} - 5\gamma = \dfrac{71}{12} - 5\gamma$

$\displaystyle \int_0^1 \dfrac{t^6}{(1-t)^2} \, dt = \int_0^1 \dfrac{(1-t)^6}{t^2} \, dt = -\dfrac{1}{2} + H_5^{(2)} - 6\gamma = \dfrac{41}{5} - 6\gamma$

Trigonometric Integrals

$\displaystyle \int_0^{\infty_+} \sin t \, dt = 1 - \cos1$

$\displaystyle \int_0^{\infty_+} \cos t \, dt = -\sin1$

$\displaystyle \int_0^{\infty_+} \tan t \, dt = -\ln(\cos 1)$

$\displaystyle \int_0^{\infty_+} \sin 2t \, dt = \dfrac{1 - \cos2}{2}$

$\displaystyle \int_0^{\infty_+} \cos 2t \, dt = -\dfrac{\sin2}{2}$

$\displaystyle \int_0^{\infty_+} \tan 2t \, dt = -\dfrac{\ln(\cos2)}{2}$

$\displaystyle \int_0^{\infty_+} \csc 2t \, dt = \dfrac{1}{2} \, \left(\ln\!\left(\tan\dfrac{1}{2}\right) + \gamma + i\pi\right)$

$\displaystyle \int_0^{\infty_+} \sec 2t \, dt = \dfrac{1}{2} \ln\!\left(\tan\!\left(\dfrac{\pi}{4}-\dfrac{1}{2}\right)\right) = \dfrac{1}{2} \ln(\sec1 - \tan1)$

$\displaystyle \int_0^{\infty_+} \cot 2t \, dt = \dfrac{1}{2} \left(\ln(\sin2) + \gamma + i\pi\right)$

$\displaystyle \int_0^{\infty_+} t\sin t \, dt = \cos1 - \sin1$

$\displaystyle \int_0^{\infty_+} t\cos t \, dt = \cos1 - 1 + \sin1$

Exponential Integrals

$\displaystyle \int_0^{\infty_+} e^t \, dt = -0! = -1$

$\displaystyle \int_0^{\infty_+} t\,e^t \, dt = 1! = 1$

$\displaystyle \int_0^{\infty_+} t^2\,e^t \, dt = -2! = -2$

$\displaystyle \int_0^{\infty_+} t^3\,e^t \, dt = 3! = 6$

exp(2πt) – 1 as the denominator

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

$\displaystyle \int_0^{\infty_+} \dfrac{e^t}{e^{2\pi t}-1} \, dt = -\dfrac{1}{2\pi}\,\Psi^{(0)}\!\left(1+\dfrac{1}{2\pi}\right) = \dfrac{1}{2\pi} \left(\gamma-H_\frac{1}{2\pi}\right)$

exp(2πt) + 1 as the denominator

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

exp(t) – 1 as the denominator

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

exp(t)+ 1 as the denominator

$\displaystyle \int_0^{\infty_+} \dfrac{1}{e^t+1} \, dt = \eta(1) = \ln 2$

Logarithmic Integrals

$\displaystyle \int_0^{\infty_+} \ln t \, dt = 0$

$\displaystyle \int_0^1 \dfrac{\ln t}{1-t} \, dt = \int_0^1 \dfrac{\ln(1-t)}{t} \, dt \\ = H_{0,2} - \zeta(2) = - \dfrac{\pi^2}{6}$

$\displaystyle \int_0^1 \dfrac{t \ln t}{1-t} \, dt = \int_0^1 \dfrac{(1-t) \ln(1-t)}{t} \, dt \\ = H_{1,2} - \zeta(2) = 1 - \dfrac{\pi^2}{6}$

$\displaystyle \int_0^1 \dfrac{t^2 \ln t}{1-t} \, dt = \int_0^1 \dfrac{(1-t)^2 \ln(1-t)}{t} \, dt \\ = H_{2,2} - \zeta(2) = \dfrac{5}{4} - \dfrac{\pi^2}{6}$

$\displaystyle \int_0^1 \dfrac{t^3 \ln t}{1-t} \, dt = \int_0^1 \dfrac{(1-t)^3 \ln(1-t)}{t} \, dt \\ = H_{3,2} - \zeta(2) = \dfrac{49}{36} - \dfrac{\pi^2}{6}$

$\displaystyle \int_0^1 \dfrac{t^4 \ln t}{1-t} \, dt = \int_0^1 \dfrac{(1-t)^4 \ln(1-t)}{t} \, dt \\ = H_{4,2} - \zeta(2) = \dfrac{205}{144} - \dfrac{\pi^2}{6}$

$\displaystyle \int_0^1 \dfrac{t^5 \ln t}{1-t} \, dt = \int_0^1 \dfrac{(1-t)^5 \ln(1-t)}{t} \, dt \\ = H_{5,2} - \zeta(2) = \dfrac{5269}{3600} - \dfrac{\pi^2}{6}$

$\displaystyle \int_0^1 \dfrac{t^6 \ln t}{1-t} \, dt = \int_0^1 \dfrac{(1-t)^6 \ln(1-t)}{t} \, dt \\ = H_{6,2} - \zeta(2) = \dfrac{5369}{3600} - \dfrac{\pi^2}{6}$

Special Function Integrals

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

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