Page:Philosophical Transactions - Volume 145.djvu/189

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170

MR. W.H.L. RUSSELL ON THE THEORY OF DEFINITE INTEGRALS.

Again, from Laplace's theorem, we have

$\int _{-\infty }^{\infty }\!\int _{-\infty }^{\infty }d\alpha dz'\cos \alpha (z=z'+x\varphi _{1}\varphi _{2}z')\varphi _{1}\varphi _{2}z'f'\varphi _{2}z'=2\pi {\frac {du}{dx}},$

where

u=f(y),\ y=\varphi _{2}(z+x\varphi _{1}y).

These theorems of course suppose the series from whence they were derived to be convergent.

As examples we may take the following.

Let	$y=1+xy^{3},$
then	$\int _{-\infty }^{\infty }\!\int _{-\infty }^{\infty }d\alpha dz'\cos \alpha (1-z+xz^{2})^{3}$

=2\pi {\frac {d}{dx}}\left\{{\sqrt[{3}]{{\Biggl (}{\frac {1}{2x}}+{\sqrt {\left({\frac {1}{4x^{2}}}-{\frac {1}{27x^{3}}}\right)}}{\Biggr )}}}+{\sqrt[{3}]{{\Biggl (}{\frac {1}{2x}}-{\sqrt {\left({\frac {1}{4x^{2}}}-{\frac {1}{27x^{3}}}\right)}}{\Biggr )}}}\right\}.

Also let	$y=1+x\varepsilon ^{y},$
then	$\int _{-\infty }^{\infty }\!\int _{-\infty }^{\infty }\cos \alpha (1-z'+x\varepsilon ^{z'})\varepsilon ^{z}d\alpha dz={\frac {2\pi \varepsilon ^{y}}{1-x\varepsilon ^{y}}},$

which we may modify thus; by eliminating $(x)$

\int _{-\infty }^{\infty }\!\int _{-\infty }^{\infty }\cos \alpha \left({\frac {(y-1\varepsilon ^{z}-(z-1)\varepsilon ^{y}}{\varepsilon ^{y}}}\right)\varepsilon ^{z}d\alpha dz={\frac {2\pi \varepsilon ^{y}}{2-y}}.

Analogous methods apply to series involving Bernouilli's numbers; thus we have

{\begin{aligned}&{\frac {x}{\varepsilon ^{x}-1}}=1-{\frac {x}{2}}+{\frac {\mathrm {B_{1}} }{1.2}}x^{2}-{\frac {\mathrm {B_{3}} }{1.2.3.4}}\cdot x^{4}+\mathrm {\&c.} \\&{\frac {\mathrm {B_{2n-1}} }{\Gamma (2n+1)}}={\frac {1}{2^{2n-1}\pi ^{2n}}}\left({\frac {1}{1^{2n}}}+{\frac {1}{2^{2n}}}+{\frac {1}{3^{2n}}}+\mathrm {\&c.} \right)\\&={\frac {1}{2^{2n-1}\pi ^{2n}\Gamma 2n}}\int _{0}^{1}{\frac {\left(\log _{\varepsilon }{\frac {1}{z}}\right)^{2n-1}dz}{1-z}};\\\therefore &\ {\frac {x}{\varepsilon ^{x}-1}}+{\frac {x}{2}}-1={\frac {x}{\pi }}\int _{0}^{1}{\frac {dz\sin \left({\frac {x}{2\pi }}\log _{\varepsilon }{\frac {1}{z}}\right)}{1-z}}.\end{aligned}}

Hence we have

\int _{0}^{1}{\frac {\sin(\alpha \log _{\varepsilon }z)dz}{z-1}}={\frac {\pi }{2}}\cdot {\frac {\varepsilon ^{2\alpha \pi }+1}{\varepsilon ^{2\alpha \pi }-1}}-{\frac {1}{2\alpha }}.

In this formula ${\textstyle (\alpha )}$ must lie between 0 and 1, as it is necessary for the convergence of the above series that $x$ should be less than $2\pi$ .

I now enter upon the consideration of the processes I have before mentioned for reducing multiple integrals to single ones. We easily see the truth of the following equation:—

{\begin{aligned}&1+{\frac {\mu }{{\frac {1}{2}}.1^{2}.2^{2}}}+{\frac {\mu ^{2}}{{\frac {1}{2}}\cdot {\frac {3}{2}}.1^{2}.2^{2}.2^{4}}}+{\frac {\mu ^{3}}{{\frac {1}{2}}\cdot {\frac {3}{2}}\cdot {\frac {5}{2}}.1^{2}.2^{2}.3^{2}.2^{6}}}+\mathrm {\&c.} \\&=2+{\frac {\mu }{1.2.1}}+{\frac {\mu ^{2}}{1.2.3.4.1.2}}+{\frac {\mu ^{3}}{1.2.3.4.5.6.1.2.3}}+\mathrm {\&c.} -1.\end{aligned}}