Pagina:Werkecarlf03gausrich.djvu/165

si $m<\mu ,$ per eliminationem erui potest.^[1] Quin adeo semissis talium integralium sufficiet, si formulam 54 simul in auxilium vocamus. Ita e.g. statuendo

$\int {\frac {\operatorname {d} x}{\sqrt[{5}]{(1-x^{5})}}}=C,\quad$$\int {\frac {\operatorname {d} x}{\sqrt[{5}]{(1-x^{5})^{2}}}}=D,\quad$$\int {\frac {\operatorname {d} x}{\sqrt[{5}]{(1-x^{5})^{3}}}}=E,\quad$$\int {\frac {\operatorname {d} x}{\sqrt[{5}]{(1-x^{5})^{4}}}}=F,$

erit

$C=\Pi {\tfrac {1}{5}}.\Pi (-{\tfrac {1}{5}}),\quad$$D={\frac {\Pi {\tfrac {1}{5}}.\Pi (-{\tfrac {2}{5}})}{\Pi (-{\frac {1}{5}})}},\quad$$E={\frac {\Pi {\tfrac {1}{5}}.\Pi (-{\tfrac {3}{5}})}{\Pi (-{\tfrac {2}{5}})}},\quad$$F={\frac {\Pi {\tfrac {1}{5}}.\Pi {(-{\tfrac {4}{5}})}}{\Pi (-{\frac {3}{5}})}}$

Hinc propter $\Pi {\tfrac {1}{5}}={\tfrac {1}{5}}\Pi (-{\tfrac {4}{5}}),$ habemus

$\Pi (-{\tfrac {1}{5}})={\sqrt[{5}]{\frac {5C^{4}}{DEF}}},\quad$$\Pi (-{\tfrac {2}{5}})={\sqrt[{5}]{\frac {25C^{3}D^{3}}{EEFF}}},\quad$$\Pi (-{\tfrac {3}{5}})={\sqrt[{5}]{\frac {125CCDDEE}{F^{3}}}},\quad$$\Pi (-{\tfrac {4}{5}})={\sqrt[{5}]{625CDEF}}$

Formulae 54, 55 adhuc suppeditant

$C={\frac {\pi }{\sin {\tfrac {1}{5}}\pi }},\quad$${\frac {D}{F}}={\frac {\sin {\tfrac {1}{5}}\pi }{\sin {\tfrac {2}{5}}\pi }}$

ita ut duo integralia $D,E,$ vel $E$ et ${F}$ sufficiant, ad omnes valores $\Pi (-{\tfrac {1}{5}}),$ $\Pi (-{\tfrac {2}{5}})$ etc. computandos.

28.

Statuendo $y=\nu x,$ atque $\mu =1,$ ${\tfrac {\Pi \lambda .\Pi \nu }{\lambda \Pi (\lambda +\nu )}}$ erit valor integralis $\int {\frac {y^{2-1}(1-{\frac {y}{\nu }})^{\nu }\operatorname {d} y}{\nu ^{2}}}$ ab $y=0$ usque ad $y=\nu ,$ sive valor integralis $\int y^{\lambda -1}(1-{\tfrac {y}{\nu }})^{\nu }\operatorname {d} y$ inter eosdem limites $={\frac {\nu ^{\lambda }\Pi \lambda .\Pi \nu }{\lambda \Pi (\lambda +\nu )}}={\frac {\Pi (\nu ,\lambda )}{\lambda }}$ (form. 47), siquidem $\nu$ denotet integrum. Iam crescente $\nu$ in infinitum, limes ipsius $\Pi (\nu ,\lambda )$ erit $=\Pi \lambda ,$ limes ipsius $(1-{\tfrac {y}{\nu }})^{\nu }$ autem $e^{-y},$ denotante $e$ basin logarithmorum hyperbolicorum. Quamobrem si $\lambda$ est positiva, ${\tfrac {\Pi \lambda }{\lambda }}$ sive $\Pi (\lambda -1)$ exprimet integrale $\int y^{\lambda -1}e^{-y}\operatorname {d} y$ ab $y=0$ usque ad $y=\infty ,$ sive scribendo $\lambda$ pro $\lambda -1,$ $\Pi \lambda$ est valor integralis $\int y^{\lambda }e^{-y}\operatorname {d} y$ ab $y=0$ usque ad $y=\infty ,$ si $\lambda +1$ est quantitas positiva.

Generalius statuendo $y=z^{\alpha },$ $\alpha \lambda +\alpha -1=\beta ,$ transit $\int y^{\lambda }e^{-y}\operatorname {d} y$ in $\int \alpha z^{\beta }e^{-z^{\alpha }}\operatorname {d} z,$ quod itaque inter limites $z=0$ atque $z=\infty$ sumtum exprimetur per $\Pi ({\tfrac {\beta +1}{\alpha }}-1)$ sive

Valor integralis $\int z^{\beta }e^{-z^{\alpha }}\operatorname {d} z,$ a $z=0$ usque ad $z=\infty$ fit $={\frac {\Pi ({\frac {\beta +1}{\alpha }}-1)}{\alpha }}={\frac {\Pi {\frac {\beta +1}{\alpha }}}{\beta +1}}$ si modo $\alpha$ atque $\beta +1$ sunt quantitates positivae (si utraque est negativa, in-

1. Haec eliminatio, si pro quantitatibus ipsis logarithmos introducimus, aequationibus tantammodo linearibus applicanda erit.