# Page:Philosophical Transactions - Volume 145.djvu/187

168
mr. w.h.l. russell on the theory of definite integrals.

equations in finite terms would be practicable in very few cases. The following method of determining a well-known definite integral is here added, to show the connexion between previous investigations relative to definite integrals, and those given in the present memoir.

 We know that ${\displaystyle 1-r^{2}+{\frac {r^{4}}{1.2}}-{\frac {r^{6}}{1.2.3}}+\dots =\varepsilon ^{-r^{2}},}$
 or ${\displaystyle 1-{\frac {(2r)^{2}}{1.2}}\cdot {\frac {1}{2}}+{\frac {(2r)^{4}}{1.2.3.4}}\cdot {\frac {1.3}{2^{2}}}-{\frac {(2r)^{6}}{1.2.3.4.5.6}}\cdot {\frac {1.3.5}{2^{3}}}+\mathrm {\&c.} =\varepsilon ^{-r^{2}}.}$
 Hence remembering that ${\displaystyle \int _{0}^{\infty }dz\ z^{2n}\varepsilon ^{-z^{2}}={\frac {1.3..2n-1}{2^{n}}}\cdot {\frac {\sqrt {\pi }}{2}},}$
 we find ${\displaystyle \int _{0}^{\infty }\varepsilon ^{-z^{2}}\cos 2rz={\frac {\sqrt {\pi }}{2}}\varepsilon ^{-r^{2}}.}$

I shall now enter on some investigations connected with Lagrange's theorem.

Let ${\displaystyle 1-y+\alpha y^{r}=0}$ be an algebraical equation. Then Lagrange's theorem gives us the following series:—

${\displaystyle y^{m}=1+m\alpha +{\frac {m(m+2r-1)}{1.2}}\alpha ^{2}+\mathrm {\&c.} +{\frac {m(m+nr-1)(m+nr-2).\ldots (m+n(r-1)+1)}{1.2.3\ldots n}}\alpha ^{n}+\mathrm {\&c.} }$

If we apply the usual test of convergency to this series, we find that ${\displaystyle (r-1)\alpha }$ must be less than unity.

Then we see that

{\displaystyle {\begin{aligned}{\frac {1}{m}}\cdot {\frac {dy^{m}}{d\alpha }}&=1+(m+2r-1)\alpha +{\frac {(m+3r-1)(m+3r-2)}{1.2}}\alpha ^{2}+\mathrm {\&c.} \\&+{\frac {(m+nr-1)(m+nr-2)\ldots (m+n(r-1)+1)}{1.2.3\ldots (n-1)}}\alpha ^{n-1}+\mathrm {\&c.} \end{aligned}}}

 Now ${\displaystyle (m+nr-1)(m+nr-2)\ldots (m+n(r-1)+1)={\frac {\Gamma (m+nr)}{\Gamma (m+n(r-1)+1)}}}$
 wherefore, since ${\displaystyle {\frac {\Gamma (a+b-1)}{\Gamma a\Gamma b}}={\frac {2^{a+b-2}}{\pi }}\int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}\cos ^{a+b-2}\theta \ \varepsilon ^{(a-b)i\theta }d\theta ,}$
 we have ${\displaystyle (m+nr-1)\ldots (m+n(r-1)+1)}$

{\displaystyle {\begin{aligned}&={\frac {2^{m+nr-1}\Gamma (n)}{\pi }}\int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}\cos ^{m+nr-1}\theta \ \varepsilon ^{(m+n(r-2)+1)i\theta }d\theta \\&={\frac {2^{m+nr-1}}{\pi }}\int _{0}^{\infty }\int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}d\theta dz\cos ^{m+nr-1}\theta \ \varepsilon ^{(m+n(r-2)+1)i\theta }.z^{n-1}\varepsilon ^{-z}.\end{aligned}}}

 Hence we have ${\displaystyle 1+(m+2r-1)\alpha +{\frac {(m+3r-1)(m+3r-2)}{1.2}}\alpha ^{2}+\mathrm {\&c.} }$

{\displaystyle {\begin{aligned}&={\frac {2^{m+r-1}}{\pi }}\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 );\end{aligned}}}