Page:Philosophical Transactions - Volume 145.djvu/182

The following are instances of the application of this method obtained by using series I., III., IV.:—

${\begin{aligned}\int _{0}^{1}\int _{-\pi }^{\pi }&d\theta dz(1-z)^{\frac {1}{2}}\varepsilon ^{\mu (\alpha +z)\cos \theta }\cos(a\mu \sin \theta )\cos(\mu z\sin \theta )\\&={\frac {2\pi }{3}}+{\frac {\pi }{8\mu ^{3}\alpha ^{\frac {3}{2}}}}\left\{(2\mu {\sqrt {\alpha }}-1)\varepsilon ^{2\mu {\sqrt {\alpha }}}+(2\mu {\sqrt {\alpha }}+1)\varepsilon ^{-2\mu {\sqrt {\alpha }}}\right\}\end{aligned}}$

${\begin{aligned}\int _{0}^{1}\int _{0}^{1}&\int _{-\pi }^{\pi }\int _{-\pi }^{\pi }\varepsilon ^{\alpha \cos \theta +\beta \cos \varphi +\mu vz\cos(\theta +\varphi )}(1-v)^{\frac {1}{3}}(1-z)^{\frac {2}{3}}\\&\cos(\alpha \sin \theta )\cos(\beta \sin \varphi )\cos \left(\mu vz\sin(\theta +\varphi )\right)d\theta d\varphi dvdz\\&={\frac {27\pi ^{2}}{20}}+{\frac {2\pi ^{2}}{81(\mu \alpha \beta )^{\frac {5}{3}}}}(3{\sqrt[{3}]{\mu \alpha \beta }}-2)\varepsilon {3{\sqrt[{3}]{\mu \alpha \beta }}}-{\frac {2\pi ^{2}}{81(\mu \alpha \beta )^{\frac {5}{3}}}}\varepsilon ^{\frac {-3{\sqrt[{3}]{\mu \alpha \beta }}}{2}}\\&\left\{(3{\sqrt[{3}]{\mu \alpha \beta }}-2)\cos {\frac {3{\sqrt[{3}]{\mu \alpha \beta }}}{2}}{\sqrt[{3}]{\mu \alpha \beta }}-{\sqrt {3}}(3{\sqrt[{3}]{\mu \alpha \beta }}+2)\sin {\frac {3{\sqrt[{3}]{\mu \alpha \beta }}}{2}}{\sqrt[{3}]{\mu \alpha \beta }}\right\}\end{aligned}}$

${\begin{aligned}\int _{0}^{1}\int _{-\pi }^{\pi }&d\theta dv\ v(1-v)^{-{\frac {1}{2}}}\varepsilon ^{(\alpha +\mu v)\cos \theta }\cos(2\theta +\mu v\sin \theta )\cos(\alpha \sin \theta )\\&={\frac {\pi }{\mu ^{2}}}+{\frac {\pi }{2\mu ^{2}}}({\sqrt {\alpha \mu }}-1)\varepsilon ^{2{\sqrt {\alpha \mu }}}-{\frac {\pi }{2\mu ^{2}}}({\sqrt {\alpha \mu }}+1)\varepsilon ^{-2{\sqrt {\alpha \mu }}}.\end{aligned}}$

Again, we know that

$\int _{0}^{\frac {\pi }{2}}d\theta \cos ^{\beta }\theta \cos n\theta ={\frac {\pi \Gamma (\beta +1)}{2^{\beta +1}\Gamma \left({\frac {\beta +n}{2}}+1\right)\Gamma \left({\frac {\beta -n}{2}}+1\right)}},$

from which we may deduce the following:

$\int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}\cos ^{a+b-2}\theta \varepsilon ^{(a-b)i\theta }d\theta ={\frac {\pi \Gamma (a+b-1)}{2^{a+b-2}\Gamma a\Gamma b}}$

Now consider the series

$1+{\frac {\alpha }{\beta }}x+{\frac {\alpha (\alpha +1)}{\beta (\beta +1)}}\cdot {\frac {x^{2}}{1.2}}+{\frac {\alpha (\alpha +1)(\alpha +2)}{\beta (\beta +1)(\beta +2)}}\cdot {\frac {x^{3}}{1.2.3}}+\mathrm {\&c.}$

where ($\alpha$) is greater than $\beta$. Then by the above formula

${\frac {\Gamma (\alpha +n)}{\Gamma (\beta +n)}}={\frac {2^{\alpha +n-1}}{\pi }}\Gamma (\alpha -\beta +1)\int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}d\theta \cos ^{\alpha +n-1}\theta \varepsilon ^{(2\beta -\alpha +n-1)i\theta };$

and we find for the sum of the series,

${\frac {2^{\alpha -1}}{\pi }}\cdot {\frac {\Gamma \beta \Gamma (\alpha -\beta +1)}{\Gamma \alpha }}\int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}d\theta \cos ^{\alpha -1}\theta \ \varepsilon ^{(2\beta -\alpha -1)i\theta }\varepsilon ^{2\cos \varepsilon ^{i\theta }x}.$

In like manner we can find the sum of the series

$1+{\frac {\alpha }{\beta }}\cdot {\frac {\alpha '}{\beta '}}x+{\frac {\alpha (\alpha +1)}{\beta (\beta +1)}}\cdot {\frac {\alpha '(\alpha '+1)}{\beta '(\beta '+1)}}\cdot {\frac {x^{2}}{1.2}}+\mathrm {\&c.,}$

where $\alpha$ is greater than $\beta$, $\alpha '$ than $\beta '$.