Convergent Integrals

TRIBE

Laplace and Mellin Transforms (r > 0 and s > 0 and u > 0)

$\displaystyle \int_0^{\infty_+} t^{s-1} \, e^{-rt} \, dt = \dfrac{\Gamma(s)}{r^s}$

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

$\displaystyle \int_0^{\infty_+} t^{s-1} \, \dfrac{e^{-rt}}{1-e^{-ut}} \, dt = \int_0^{\infty_+} t^{s-1} \, \dfrac{e^{-(r-u)t}}{e^{ut}-1} \, dt = \dfrac{\Gamma(s)}{u^s} \, \zeta\!\left(s,\frac{r}{u}\right)$

$\displaystyle \int_0^{\infty_+} t^{s-1} \, e^{-rt} \, \ln t \, dt = \dfrac{\Gamma(s)}{r^s} \, (\Psi^{(0)}(s)-\ln r) \\ \\ = \dfrac{\Gamma(s)}{r^s} \, (H_{s-1}-\gamma-\ln r)$

$\displaystyle \int_0^{\infty_+} t^{s-1} \, e^{-rt} \, (\ln t)^2 \, dt = \dfrac{\Gamma(s)}{r^s} \, \left((\Psi^{(0)}(s)-\ln r)^2 + \Psi^{(1)}(s)\right) \\ \\ = \dfrac{\Gamma(s)}{r^s} \, \left((H_{s-1}-\gamma-\ln r)^2 + \Psi^{(1)}(s)\right)$

Laplace Transforms (r > 0 and s > 0)

$\displaystyle \int_0^{\infty_+} t^n \, e^{-rt} \, dt = \dfrac{\Gamma(n\!+\!1)}{r^{n+1}} = \dfrac{n!}{r^{n+1}} \qquad (n\in\mathbb{N})$

$\displaystyle \int_0^{\infty_+} t^s \, e^{-rt} \, dt = \dfrac{\Gamma(s\!+\!1)}{r^{s+1}}$

$\displaystyle \int_0^{\infty_+} \dfrac{e^{-rt}}{t^{n+1}} \, dt = (-1)^n \, \dfrac{r^n}{n!} \, (H_n - \ln r) \qquad (n\in\mathbb{N})$

$\displaystyle \int_0^{\infty_+} \dfrac{e^{-rt}}{t^{s+1}} \, dt = \dfrac{(-r)^s}{\Gamma(s\!+\!1)} \, (H_s - \ln r)$

Mellin Transforms (r > 0 and s > 0)

exp(2πt) – 1 as the denominator

$\displaystyle i\int_0^{\infty_+} \dfrac{(it)^r\,(e^{-iat}-(-1)^r\,e^{iat})}{e^{2\pi t}-1} \, dt = \text{Li}_{-r}(e^{-a}) \,-\, \dfrac{\Gamma(r\!+\!1)}{a^{r+1}}$

$\displaystyle \int_0^{\infty_+} \dfrac{t^{2n}\sin(at)}{e^{2\pi t}-1} \, dt = \dfrac{(-1)^r}{2} \left(\text{Li}_{-2n}(e^{-a}) \,-\, \dfrac{\Gamma(2n\!+\!1)}{a^{2n+1}}\right)$

$\displaystyle \int_0^{\infty_+} \dfrac{t^{2n-1}\cos(at)}{e^{2\pi t}-1} \, dt = \dfrac{(-1)^r}{2} \left(\text{Li}_{1-2n}(e^{-a}) \,-\, \dfrac{\Gamma(2n)}{a^{2n}}\right)$

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