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#### chisigma

##### Well-known member

- Feb 13, 2012

- 1,704

Integral Caculation

... the user widapol did have some difficulties in the computation of the integral...

$\displaystyle \int_{0}^{\infty} \ln^{2} (1+t)\ e^{- s t}\ dt$ (1)

... which of course is the L-Transform of the function $\displaystyle \ln^{2} (1+t)$. Remembering thye basic property of the L-Transform...

$\displaystyle \mathcal{L}\{f(t+a)\} = e^{-a\ s}\ \mathcal{L}\{f(t)\}$ (2)

... and that...

$\displaystyle \mathcal{L}\{\ln^{2} t\} = \frac{\frac{\pi^{2}}{6} + (\gamma + \ln s)^{2}}{s}$ (3)

... where $\displaystyle \gamma=.5772156...$ is the Euler's constant, we easily find...

$\displaystyle \int_{0}^{\infty} \ln^{2} (1+t)\ e^{- s t}\ dt = e^{-s}\ \frac{\frac{\pi^{2}}{6} + (\gamma + \ln s)^{2}}{s}$ (4)

A little less easy is to demonstrate that (3) is true... that will be done in a successive post...

Kind regards

$\chi$ $\sigma$