# Page:Philosophical Transactions - Volume 145.djvu/188

169
mr. w.h.l. russell on the theory of definite integrals.

{\displaystyle {\begin{aligned}\therefore \ \int _{0}^{\infty }\!&\int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}d\theta dz\cos ^{m+r-1}\theta \ \varepsilon ^{2^{r}az\cos ^{r}\theta \cos(r-2)\theta -z}\\&\cos(2^{r}az\cos ^{r}\theta \sin(r-2)\theta +(m+r-1)\theta )={\frac {\pi }{2^{m+r-1}m}}\cdot {\frac {d.y^{m}}{da}}\end{aligned}}}

Let ${\displaystyle r=2}$, then we have

${\displaystyle \int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}{\frac {d\theta \cos ^{m+1}\theta \cos(m+1)\theta }{1-c\cos ^{2}\theta }}={\frac {2\pi }{m}}\cdot {\frac {d}{dc}}\left\{{\frac {1-{\sqrt {1-c}}}{c}}\right\}^{m},}$

where ${\displaystyle (c)}$ is of course less than unity; an integral given by Abel.

When ${\displaystyle 2^{r}\alpha }$ is less than unity we can always integrate with respect to ${\displaystyle (z)}$, but may obtain a single integral more simply by proceeding as follows:—

 We have ${\displaystyle {\frac {(m+nr-1)(m+nr-2)\ldots (m+n(r-1)+1)}{1.2.3\ldots n-1}}}$

${\displaystyle ={\frac {2^{m}+nr-1}{\pi }}\int _{={\frac {\pi }{2}}}^{\frac {\pi }{2}}d\theta \cos ^{m+nr-1}\theta \ \varepsilon ^{(m+n_{r}-2)+1)i\theta };}$

consequently we find by summing a geometrical progression,

${\displaystyle \int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}d\theta \cos ^{m+r-1}\theta \left\{{\frac {\cos(m+r-1)\theta -2^{r}a\cos ^{r}\theta \cos(m+1)\theta }{1-2^{r+1}a\cos ^{r}\theta \cos(r-2)\theta +2^{2r}a^{2}\cos ^{2r}\theta }}\right\}={\frac {\pi }{2^{m+r-1}m}}{\frac {dy^{m}}{da}}.}$

When ${\displaystyle r=2}$ this result coincides with that last obtained. We may obtain a very general result by applying Fourier's theorem to the series of Lagrange and Laplace as follows:—

If ${\displaystyle u=f(y)}$, and ${\displaystyle y=z+x\varphi (y)}$,

 we have ${\displaystyle u=f(z)+\{\varphi (z)f'(z)\}x+{\frac {d}{dz}}\{\varphi ^{2}zf'z\}{\frac {x^{2}}{1.2}}+\mathrm {\&c,;} }$

${\displaystyle \therefore \ {\frac {du}{dx}}=\varphi (z)f'(z)+{\frac {d}{dz}}\{\varphi ^{2}(z)f'z\}x+{\frac {d^{2}}{dz^{2}}}\{\varphi ^{3}(z)f'(z)\}{\frac {x^{2}}{1.2}}+\mathrm {\&c.;} }$

 Now we generally have ${\displaystyle \mathrm {F} (z)=\int _{-\infty }^{\infty }\!\int _{-\infty }^{\infty }\cos \alpha (z-z')\mathrm {F} z'{\frac {dx.dz'}{2\pi }},}$ whence ${\displaystyle \varphi ^{n}(z)f'z=\int _{-\infty }^{\infty }\!\int _{-\infty }^{\infty }\varepsilon ^{i\alpha (z-z')}\varphi ^{n}(z')f'(z'){\frac {d\alpha .dz'}{2\pi }}}$ and ${\displaystyle {\frac {d^{n-1}}{dz^{n-1}}}\varphi ^{n}(z)f'(z)=\int _{-\infty }^{\infty }\!\int _{-\infty }^{\infty }\varepsilon ^{i\alpha (z-z')}(i\alpha )^{n-1}\varphi ^{n}(z')f'(z'){\frac {d\alpha .dz'}{2\pi }}.}$

Hence substituting in the above series, we find

${\displaystyle {\frac {du}{dx}}=\int _{-\infty }^{\infty }\!\int _{-\infty }^{\infty }\varepsilon ^{i\alpha (z-z')}\varphi (z')f'(z')\varepsilon ^{i\alpha \varphi (z')x}{\frac {d\alpha dz'}{2\pi }}.}$

Consequently we find the following definite integral:

${\displaystyle \int _{-\infty }^{\infty }\!\int _{-\infty }^{\infty }d\alpha dz\varphi (z')f'(z')\cos \alpha {\bigl (}z-z'+x\varphi (z'){\bigr )}=2\pi {\frac {du}{dx}}.}$