Page:Philosophical Transactions - Volume 145.djvu/183

The use of this integral will give an important extension of the method I have employed for expressing the integrals of differential equations by means of definite integrals. For in order to the success of that method, it is necessary, as is shown in my paper in the Cambridge Mathematical Journal before alluded to, that the magnitude of the factorials (if any) in the numerator of each term of the series to be summed, should be less than that of the corresponding factorials in the denominator; whereas this integral enables us to sum series in which the reverse is the case. I shall now apply the series, whose sum we have just found, to the evaluation of definite integrals, using series VI. and VII. Hence

${\begin{aligned}&\int _{0}^{1}\int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}d\theta dv\ v(1-v)^{2}\cos ^{3}\theta \varepsilon ^{2\mu v\cos ^{2}\!\theta }\cos(2\mu v\sin \theta \cos \theta +\theta )={\frac {\pi }{4\mu ^{4}}}(\mu ^{2}-3\mu +3)\varepsilon ^{\mu }-{\frac {3\pi }{4\mu ^{4}}}+{\frac {\pi }{8\mu ^{2}}}\\&\int _{0}^{1}\int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}d\theta dv\ v(1-v)^{3}\cos ^{4}\theta \varepsilon ^{2\mu v\cos ^{2}\!\theta }\cos(2\mu v\sin \theta \cos \theta +2\theta )\end{aligned}}$

$={\frac {3\pi }{8\mu ^{5}}}(\mu -2)^{2}\varepsilon ^{\mu }+{\frac {\pi }{8\mu ^{3}}}(\mu +3)-{\frac {3\pi }{2\mu ^{5}}}.$

${\displaystyle }$

By a process similar to those used above, we find

$1+{\frac {2}{{\frac {3}{2}}.3.1}}{\frac {\mu ^{2}x^{2}}{2^{2}}}+{\frac {2.3}{{\frac {3}{2}}\cdot {\frac {5}{2}}.3.4.1.2}}{\frac {\mu ^{4}x^{4}}{2^{4}}}+\mathrm {\&c.}$

$=-{\frac {8}{\mu ^{4}x^{4}}}+{\frac {2}{\mu ^{4}x^{4}}}(\mu ^{2}x^{2}-2\mu x+2)\varepsilon ^{\mu x}+{\frac {2}{\mu ^{4}x^{4}}}(\mu ^{2}x^{2}+2\mu x+2)\varepsilon ^{-\mu x}.$

Hence

$\int _{-\pi }^{\pi }\int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}d\theta d\varphi \ \varepsilon ^{\alpha \cos \varphi +2\mu \cos \theta \cos(\theta +\varphi )}\cos \theta \cos(\alpha \sin \varphi )\cos 2\left(\varphi +\mu \cos \theta \sin(\theta +\varphi )\right)$

$=-{\frac {\pi }{2\mu ^{2}}}+{\frac {\mu }{4\mu ^{2}}}(2\mu \alpha -2{\sqrt {\mu \alpha }}+1)\varepsilon ^{2{\sqrt {\mu \alpha }}}+{\frac {\pi }{4\mu ^{2}}}(2\mu \alpha +2{\sqrt {\mu \alpha }}+1)\varepsilon ^{-2{\sqrt {\mu \alpha }}}.$

The following formulæ are found in Crelle's Journal:—

${\begin{aligned}&\int _{0}^{\frac {\pi }{2}}\cos ^{a-2}\theta \cot ^{b}\theta \cos a\theta d\theta ={\frac {\Gamma (a+b-1)}{\Gamma (a)\Gamma (b)}}\cdot {\frac {\pi }{2\cos {\frac {b\pi }{2}}}}\\&\int _{0}^{\frac {\pi }{2}}\cos ^{a-2}\theta \cot ^{b}\theta \sin a\theta d\theta ={\frac {\Gamma (a+b-1)}{\Gamma (a)\Gamma (b)}}\cdot {\frac {\pi }{2\sin {\frac {b\pi }{2}}}};\\&\therefore \ \int _{0}^{\frac {\pi }{2}}\cos ^{a-2}\theta \cot ^{b}\theta \ \varepsilon ^{ai\theta }d\theta ={\frac {\Gamma (a+b-1)}{\Gamma a\Gamma b}}\cdot {\frac {\pi }{\sin b\pi }}\varepsilon ^{i(1-b){\frac {\pi }{2}}},\end{aligned}}$

whence we find

${\frac {\Gamma (a+b-1)}{\Gamma a}}={\frac {\Gamma b\sin b\pi }{\pi }}\varepsilon ^{-i(1-b){\frac {\pi }{2}}}\int _{0}^{\frac {\pi }{2}}\cos ^{a-2}\theta \cot ^{b}\theta \ \varepsilon ^{ai\theta }d\theta .$

In this formula we suppose (b) to be less than unity.