A Laplace Transform...

chisigma

Well-known member
On MHF...

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

$\chi$ $\sigma$

chisigma

Well-known member
The result of the previous post would be incomplete without the computation of the integral...

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

Setting $s\ t= \xi$ the integral becomes...

$\displaystyle I(s)= \frac{1}{s}\ \int_{0}^{\infty} \ln^{2} \frac{\xi}{s}\ e^{-\xi}\ d \xi = \frac{1}{s}\ (\int_{0}^{\infty} \ln^{2} \xi\ e^{- \xi}\ d \xi - 2\ \ln s\ \int_{0}^{\infty} \ln \xi\ e^{- \xi}\ d \xi + \ln^{2} s\ \int_{0}^{\infty} e^{- \xi}\ d \xi)$ (2)

As You can see the result is the sum of three integrals. Starting with the last it is well known that...

$\displaystyle \int_{0}^{\infty} e^{- \xi}\ d \xi=1$ (3)

A little less known is the second integral that is...

$\displaystyle \int_{0}^{\infty} \ln \xi\ e^{- \xi}\ d \xi = - \gamma$ (4)

... where $\gamma= .5772156...$ is the Euler's constant.

The first integral can be written as...

$\displaystyle \int_{0}^{\infty} \ln^{2} \xi\ e^{- \xi}\ d \xi = \varphi^{\ ''} (0)$ (5)

... where...

$\displaystyle \varphi(x)= x!= \int_{0}^{\infty} t^{x}\ e^{- t}\ dt$ (6)

In...

http://www.mathhelpboards.com/f15/difference-equation-tutorial-draft-part-i-426/#post2494

... it has been obtained a series expansion for...

$\displaystyle \alpha(x)= \frac{d}{d x} \ln \varphi(x)= \frac{\varphi^{\ '} (x)}{\varphi(x)} = - \gamma + \sum_{k=2}^{\infty} (-1)^{k}\ \zeta(k)\ x^{k-1}$ (7)

... where $\zeta(*)$ is the 'Riemann Zeta Function'. Deriving (7) we obtain an explicit expression of $\varphi^{\ ''}(0)$...

$\displaystyle \alpha^{\ '}(0)= \frac{\varphi^{\ ''}(0)}{\varphi(0)} - \frac{\varphi^{\ '\ 2}(0)}{\varphi^{2}(0)} \implies \varphi^{\ ''}(0)= \alpha^{\ '}(0) + \varphi^{\ '\ 2}(0) = \zeta(2) + \gamma^{2}= \frac{\pi^{2}}{6} + \gamma^{2}$ (8)

... so that is...

$\displaystyle I(s)= \frac{1}{s}\ \{ \frac{\pi^{2}}{6}+ (\gamma+ \ln s)^{2}\}$ (9)

Kind regards

$\chi$ $\sigma$