170
MR. W.H.L. RUSSELL ON THE THEORY OF DEFINITE INTEGRALS.
Again, from Laplace's theorem, we have
∫
−
∞
∞
∫
−
∞
∞
d
α
d
z
′
cos
α
(
z
=
z
′
+
x
φ
1
φ
2
z
′
)
φ
1
φ
2
z
′
f
′
φ
2
z
′
=
2
π
d
u
d
x
,
{\displaystyle \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
=
φ
2
(
z
+
x
φ
1
y
)
.
{\displaystyle 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
+
x
y
3
,
{\displaystyle y=1+xy^{3},}
then
∫
−
∞
∞
∫
−
∞
∞
d
α
d
z
′
cos
α
(
1
−
z
+
x
z
2
)
3
{\displaystyle \int _{-\infty }^{\infty }\!\int _{-\infty }^{\infty }d\alpha dz'\cos \alpha (1-z+xz^{2})^{3}}
=
2
π
d
d
x
{
(
1
2
x
+
(
1
4
x
2
−
1
27
x
3
)
)
3
+
(
1
2
x
−
(
1
4
x
2
−
1
27
x
3
)
)
3
}
.
{\displaystyle =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
ε
y
,
{\displaystyle y=1+x\varepsilon ^{y},}
then
∫
−
∞
∞
∫
−
∞
∞
cos
α
(
1
−
z
′
+
x
ε
z
′
)
ε
z
d
α
d
z
=
2
π
ε
y
1
−
x
ε
y
,
{\displaystyle \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
)
{\displaystyle (x)}
∫
−
∞
∞
∫
−
∞
∞
cos
α
(
(
y
−
1
ε
z
−
(
z
−
1
)
ε
y
ε
y
)
ε
z
d
α
d
z
=
2
π
ε
y
2
−
y
.
{\displaystyle \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
x
ε
x
−
1
=
1
−
x
2
+
B
1
1.2
x
2
−
B
3
1.2.3.4
⋅
x
4
+
&
c
.
B
2
n
−
1
Γ
(
2
n
+
1
)
=
1
2
2
n
−
1
π
2
n
(
1
1
2
n
+
1
2
2
n
+
1
3
2
n
+
&
c
.
)
=
1
2
2
n
−
1
π
2
n
Γ
2
n
∫
0
1
(
log
ε
1
z
)
2
n
−
1
d
z
1
−
z
;
∴
x
ε
x
−
1
+
x
2
−
1
=
x
π
∫
0
1
d
z
sin
(
x
2
π
log
ε
1
z
)
1
−
z
.
{\displaystyle {\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
∫
0
1
sin
(
α
log
ε
z
)
d
z
z
−
1
=
π
2
⋅
ε
2
α
π
+
1
ε
2
α
π
−
1
−
1
2
α
.
{\displaystyle \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
(
α
)
{\textstyle (\alpha )}
must lie between 0 and 1, as it is necessary for the convergence of the above series that
x
{\displaystyle x}
should be less than
2
π
{\displaystyle 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:—
1
+
μ
1
2
.1
2
.2
2
+
μ
2
1
2
⋅
3
2
.1
2
.2
2
.2
4
+
μ
3
1
2
⋅
3
2
⋅
5
2
.1
2
.2
2
.3
2
.2
6
+
&
c
.
=
2
+
μ
1.2.1
+
μ
2
1.2.3.4.1.2
+
μ
3
1.2.3.4.5.6.1.2.3
+
&
c
.
−
1.
{\displaystyle {\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}}}