Page:Philosophical Transactions - Volume 145.djvu/178

and we find from this the series

${\displaystyle }$

Whence we find, putting $\mu$ for ${\textstyle {\frac {\mu ^{2}x^{2}}{2^{2}}}}$,

${\begin{aligned}\int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}d\theta &{\sqrt {\cos \theta }}\ \varepsilon ^{\mu \cos ^{2}\theta }\cos \left(\mu \sin \theta \cos \theta +{\frac {5\theta }{2}}-\tan \theta \right)\\&={\frac {\sqrt {\pi }}{2\mu ^{2}\varepsilon }}\left\{2\mu \varepsilon ^{2{\sqrt {\mu }}}-{\sqrt {\mu }}\ \varepsilon ^{2{\sqrt {\mu }}}+2\mu \varepsilon ^{-2{\sqrt {\mu }}}+\varepsilon ^{-2{\sqrt {\mu }}}\right\}\end{aligned}}$

Next consider the symbolical equation

$(\mathrm {D} -1)(\mathrm {D} -3)(\mathrm {D} -5)u-\mu ^{3}\varepsilon ^{3\omega }u=0$, where $\varepsilon ^{\omega }=x$;

and assume as the transformed equation

$(\mathrm {D} -1)(\mathrm {D} -2)(\mathrm {D} -3)v-\mu ^{3}\varepsilon ^{3\omega }v=0$.

Then

$u=(\mathrm {D} -2)v,$

and

$v=\mathrm {C} _{1}x\varepsilon ^{\mu x}$

where $1$, $\alpha$, $\beta$ are the three cube roots of unity.

Hence

${\displaystyle }$

We must determine $\mathrm {C} _{1}$, $\mathrm {C} _{2}$, $\mathrm {C} _{3}$ according to the series we have to sum.

If

${\displaystyle }$

we find

Whence

$\int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}\!\int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}d\theta d\varphi \varepsilon ^{\mu \cos \theta \cos \varphi \cos(\theta +\varphi )}\cos ^{-{\frac {1}{3}}}\theta \cos ^{\frac {1}{3}}\varphi$