Divergent Integrals

SUBSPECIES

Monomial Integrals

$\displaystyle \int_1^{\infty_+} dt = -1$

$\displaystyle \int_1^{\infty_+} \sqrt{t} \, dt = -\dfrac{2}{3}$

$\displaystyle \int_2^{\infty_+} \sqrt{t} \, dt = \dfrac{2}{3} \, (1-2\sqrt{2}) = \dfrac{2}{3} - \dfrac{4\sqrt{2}}{3}$

$\displaystyle \int_3^{\infty_+} \sqrt{t} \, dt = \dfrac{2}{3} \, (2\sqrt{2}-3\sqrt{3}) = \dfrac{4\sqrt{2}}{3} - 2\sqrt{3}$

$\displaystyle \int_1^{\infty_+} t \, dt = -\dfrac{1}{2}$

$\displaystyle \int_1^{\infty_+} t^2 \, dt = -\dfrac{1}{3}$

$\displaystyle \int_1^{\infty_+} t^3 \, dt = -\dfrac{1}{4}$

$\displaystyle \int_1^{\infty_+} t^4 \, dt = -\dfrac{1}{5}$

Monomial Ratio Integrals

$\displaystyle \int_1^{\infty_+} \dfrac{1}{t} \, dt = \ln\infty_+ - \ln 1 = 0$

Polynomial Ratio Integrals

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

$\displaystyle \int_1^{\infty_+} \dfrac{1}{1-t} \, dt = \ln 0 = -\gamma$

$\displaystyle \int_\frac{3}{2}^{\infty_+} \dfrac{1}{1-t} \, dt = \ln(-1) = i\pi$

$\displaystyle \int_2^{\infty_+} \dfrac{1}{1-t} \, dt = \ln(-1) - \ln 0 = \gamma + i\pi$

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

$\displaystyle \int_1^{\infty_+} \dfrac{1}{1+t} \, dt = -\ln 2$

$\displaystyle \int_2^{\infty_+} \dfrac{1}{1+t} \, dt = \ln 2 - \ln 3$

$\displaystyle \int_1^{\infty_+} \dfrac{t}{1+t} \, dt = \ln 2 - 1$

$\displaystyle \int_1^{\infty_+} \dfrac{t^2}{1+t} \, dt = \dfrac{1}{2} - \ln 2$

$\displaystyle \int_1^{\infty_+} \dfrac{t^3}{1+t} \, dt = -\dfrac{5}{6} + \ln 2$

$\displaystyle \int_1^{\infty_+} \dfrac{t^4}{1+t} \, dt = \dfrac{7}{12} - \ln 2$

$\displaystyle \int_1^{\infty_+} \dfrac{t^5}{1+t} \, dt = -\dfrac{47}{60} + \ln 2$

$\displaystyle \int_1^{\infty_+} \dfrac{t^6}{1+t} \, dt = \dfrac{37}{60} - \ln 2$

Trigonometric Integrals

$\displaystyle \int_1^{\infty_+} \sin t \, dt = \cos1 - 1$

$\displaystyle \int_1^{\infty_+} \cos t \, dt = -\sin1$

$\displaystyle \int_1^{\infty_+} \tan t \, dt = \ln(\cos 1)$

$\displaystyle \int_1^{\infty_+} \sin 2t \, dt = \dfrac{\cos2 - 1}{2}$

$\displaystyle \int_1^{\infty_+} \cos 2t \, dt = -\dfrac{\sin2}{2}$

$\displaystyle \int_1^{\infty_+} \tan 2t \, dt = \dfrac{\ln(\cos2)}{2}$

$\displaystyle \int_1^{\infty_+} \csc 2t \, dt = \dfrac{1}{2} \left(\ln\!\left(\tan\dfrac{1}{2}\right) \!- \ln(\tan1)\right) = \dfrac{1}{2} \ln\!\left(\tan\dfrac{1}{2} \, \cot1\right)$

$\displaystyle \int_1^{\infty_+} \sec 2t \, dt = \dfrac{1}{2} \left(\ln\!\left(\tan\left(\dfrac{\pi}{4}+\dfrac{1}{2}\right)\right) - \ln\!\left(\tan\left(\dfrac{\pi}{4}+1\right)\right)\right)$