mr. w.h.l. russell on the theory of definite integrals.
169
∴
∫
0
∞
∫
−
π
2
π
2
d
θ
d
z
cos
m
+
r
−
1
θ
ε
2
r
a
z
cos
r
θ
cos
(
r
−
2
)
θ
−
z
cos
(
2
r
a
z
cos
r
θ
sin
(
r
−
2
)
θ
+
(
m
+
r
−
1
)
θ
)
=
π
2
m
+
r
−
1
m
⋅
d
.
y
m
d
a
{\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
r
=
2
{\displaystyle r=2}
, then we have
∫
−
π
2
π
2
d
θ
cos
m
+
1
θ
cos
(
m
+
1
)
θ
1
−
c
cos
2
θ
=
2
π
m
⋅
d
d
c
{
1
−
1
−
c
c
}
m
,
{\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
(
c
)
{\displaystyle (c)}
is of course less than unity; an integral given by Abel .
When
2
r
α
{\displaystyle 2^{r}\alpha }
is less than unity we can always integrate with respect to
(
z
)
{\displaystyle (z)}
, but may obtain a single integral more simply by proceeding as follows:—
We have
(
m
+
n
r
−
1
)
(
m
+
n
r
−
2
)
…
(
m
+
n
(
r
−
1
)
+
1
)
1.2.3
…
n
−
1
{\displaystyle {\frac {(m+nr-1)(m+nr-2)\ldots (m+n(r-1)+1)}{1.2.3\ldots n-1}}}
=
2
m
+
n
r
−
1
π
∫
=
π
2
π
2
d
θ
cos
m
+
n
r
−
1
θ
ε
(
m
+
n
r
−
2
)
+
1
)
i
θ
;
{\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,
∫
−
π
2
π
2
d
θ
cos
m
+
r
−
1
θ
{
cos
(
m
+
r
−
1
)
θ
−
2
r
a
cos
r
θ
cos
(
m
+
1
)
θ
1
−
2
r
+
1
a
cos
r
θ
cos
(
r
−
2
)
θ
+
2
2
r
a
2
cos
2
r
θ
}
=
π
2
m
+
r
−
1
m
d
y
m
d
a
.
{\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
r
=
2
{\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
u
=
f
(
y
)
{\displaystyle u=f(y)}
, and
y
=
z
+
x
φ
(
y
)
{\displaystyle y=z+x\varphi (y)}
,
we have
u
=
f
(
z
)
+
{
φ
(
z
)
f
′
(
z
)
}
x
+
d
d
z
{
φ
2
z
f
′
z
}
x
2
1.2
+
&
c
,
;
{\displaystyle u=f(z)+\{\varphi (z)f'(z)\}x+{\frac {d}{dz}}\{\varphi ^{2}zf'z\}{\frac {x^{2}}{1.2}}+\mathrm {\&c,;} }
∴
d
u
d
x
=
φ
(
z
)
f
′
(
z
)
+
d
d
z
{
φ
2
(
z
)
f
′
z
}
x
+
d
2
d
z
2
{
φ
3
(
z
)
f
′
(
z
)
}
x
2
1.2
+
&
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
F
(
z
)
=
∫
−
∞
∞
∫
−
∞
∞
cos
α
(
z
−
z
′
)
F
z
′
d
x
.
d
z
′
2
π
,
{\displaystyle \mathrm {F} (z)=\int _{-\infty }^{\infty }\!\int _{-\infty }^{\infty }\cos \alpha (z-z')\mathrm {F} z'{\frac {dx.dz'}{2\pi }},}
whence
φ
n
(
z
)
f
′
z
=
∫
−
∞
∞
∫
−
∞
∞
ε
i
α
(
z
−
z
′
)
φ
n
(
z
′
)
f
′
(
z
′
)
d
α
.
d
z
′
2
π
{\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
d
n
−
1
d
z
n
−
1
φ
n
(
z
)
f
′
(
z
)
=
∫
−
∞
∞
∫
−
∞
∞
ε
i
α
(
z
−
z
′
)
(
i
α
)
n
−
1
φ
n
(
z
′
)
f
′
(
z
′
)
d
α
.
d
z
′
2
π
.
{\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
d
u
d
x
=
∫
−
∞
∞
∫
−
∞
∞
ε
i
α
(
z
−
z
′
)
φ
(
z
′
)
f
′
(
z
′
)
ε
i
α
φ
(
z
′
)
x
d
α
d
z
′
2
π
.
{\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:
∫
−
∞
∞
∫
−
∞
∞
d
α
d
z
φ
(
z
′
)
f
′
(
z
′
)
cos
α
(
z
−
z
′
+
x
φ
(
z
′
)
)
=
2
π
d
u
d
x
.
{\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}}.}