Pagina:Werkecarlf03gausrich.djvu/172

Generalius statuendo $y=z^{\alpha },$ $\alpha \lambda +\alpha -1=\beta ,$ valor integralis $\int z^{\varepsilon }e^{-z^{a}}\log z\operatorname {d} z,$ a $z=0$ usque ad $z=\infty ,$ fit

$={\tfrac {1}{\alpha \alpha }}\Pi ({\tfrac {\beta +1}{\alpha }}-1).\Psi ({\tfrac {\beta +1}{\alpha }}-1)={\tfrac {1}{\alpha (\beta +1)}}\Pi {\tfrac {\beta +1}{\alpha }}.\Psi {\tfrac {\beta +1}{\alpha }}-{\tfrac {1}{(\beta +1)^{2}}}\Pi {\tfrac {\beta +1}{\alpha }}$

siquidem simul $\beta +1$ atque $\alpha$ sunt quantitates positivae, vel aequalis eidem quantitati cum signo opposito, si utraque $\beta +1,$ $\alpha$ est negativa.

35.

At non solum productum $\Pi \lambda .\Psi \lambda ,$ verum etiam ipsa functio $\Psi \lambda$ per integrale determinatum exhiberi potest. Designante $k$ integrum positivum, patet valorem integralis $\int {\frac {x^{\lambda }-x^{\lambda +k}}{1-x}}.\operatorname {d} x,$ ab $x=0$ usque ad $x=1$ esse

$={\frac {1}{\lambda +1}}+{\frac {1}{\lambda +2}}+{\frac {1}{\lambda +3}}+{\text{etc.}}+{\frac {1}{\lambda +k}}$

Porro quum valor integralis $\int \left({\frac {1}{1-x}}-{\frac {kx^{k-1}}{1-x^{k}}}\right)\operatorname {d} x$ generaliter sit $=$ Const. $+\log {\frac {1-x^{k}}{1-x}},$ idem inter limites $x=0$ atque $x=1$ erit $=\log k,$ unde patet, valorem integralis $S=\int \left({\frac {1-x^{2}+x^{2+k}}{1-x}}-{\frac {kx^{k-1}}{1-x^{k}}}\right)\operatorname {d} x$ inter eosdem limites esse

$=\log k-{\frac {1}{\lambda +1}}-{\frac {1}{\lambda +2}}-{\frac {1}{\lambda +3}}-{\text{etc.}}-{\frac {1}{\lambda +k}}$

quam expressionem denotabimus per $\Omega .$ Discerpamus integrale $S$ in duas partes

$\int \left({\frac {1-x^{\lambda }}{1-x}}\right)\operatorname {d} x+\int \left({\frac {x^{\lambda +k}}{1-x}}-{\frac {kx^{k-1}}{1-x^{k}}}\right)\operatorname {d} x$

Pars prima $\int {\frac {1-x^{\lambda }}{1-x}}.\operatorname {d} x,$ statuendo $x=y^{k}$ mutatur in

$\int {\frac {ky^{k-1}-ky^{2k+k-1}}{1-y^{k}}}\operatorname {d} y$

unde sponte patet, illius valorem ab $x=0$ usque ad $x=1,$ aequalem esse valori integralis

$\int {\frac {kx^{k-1}-kx^{\lambda k+k-1}}{1-x^{k}}}\operatorname {d} x$

inter eosdem limites, quum manifesto literam $y$ sub hac restrictione in $x$ mutare liceat. Hinc fit integrale ${S},$ inter eosdem limites

$=\int \left({\frac {x^{\lambda +k}}{1-x}}-{\frac {kx^{\lambda k+k-1}}{1-x^{k}}}\right)\operatorname {d} x$

Hoc vero integrale, statuendo $x^{k}=z,$ transit in