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
1
−
r
2
+
r
4
1.2
−
r
6
1.2.3
+
⋯
=
ε
−
r
2
,
{\displaystyle 1-r^{2}+{\frac {r^{4}}{1.2}}-{\frac {r^{6}}{1.2.3}}+\dots =\varepsilon ^{-r^{2}},}
or
1
−
(
2
r
)
2
1.2
⋅
1
2
+
(
2
r
)
4
1.2.3.4
⋅
1.3
2
2
−
(
2
r
)
6
1.2.3.4.5.6
⋅
1.3.5
2
3
+
&
c
.
=
ε
−
r
2
.
{\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
∫
0
∞
d
z
z
2
n
ε
−
z
2
=
1.3..2
n
−
1
2
n
⋅
π
2
,
{\displaystyle \int _{0}^{\infty }dz\ z^{2n}\varepsilon ^{-z^{2}}={\frac {1.3..2n-1}{2^{n}}}\cdot {\frac {\sqrt {\pi }}{2}},}
we find
∫
0
∞
ε
−
z
2
cos
2
r
z
=
π
2
ε
−
r
2
.
{\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
1
−
y
+
α
y
r
=
0
{\displaystyle 1-y+\alpha y^{r}=0}
be an algebraical equation. Then Lagrange's theorem gives us the following series:—
y
m
=
1
+
m
α
+
m
(
m
+
2
r
−
1
)
1.2
α
2
+
&
c
.
+
m
(
m
+
n
r
−
1
)
(
m
+
n
r
−
2
)
.
…
(
m
+
n
(
r
−
1
)
+
1
)
1.2.3
…
n
α
n
+
&
c
.
{\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
(
r
−
1
)
α
{\displaystyle (r-1)\alpha }
must be less than unity.
Then we see that
1
m
⋅
d
y
m
d
α
=
1
+
(
m
+
2
r
−
1
)
α
+
(
m
+
3
r
−
1
)
(
m
+
3
r
−
2
)
1.2
α
2
+
&
c
.
+
(
m
+
n
r
−
1
)
(
m
+
n
r
−
2
)
…
(
m
+
n
(
r
−
1
)
+
1
)
1.2.3
…
(
n
−
1
)
α
n
−
1
+
&
c
.
{\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
(
m
+
n
r
−
1
)
(
m
+
n
r
−
2
)
…
(
m
+
n
(
r
−
1
)
+
1
)
=
Γ
(
m
+
n
r
)
Γ
(
m
+
n
(
r
−
1
)
+
1
)
{\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
Γ
(
a
+
b
−
1
)
Γ
a
Γ
b
=
2
a
+
b
−
2
π
∫
−
π
2
π
2
cos
a
+
b
−
2
θ
ε
(
a
−
b
)
i
θ
d
θ
,
{\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
(
m
+
n
r
−
1
)
…
(
m
+
n
(
r
−
1
)
+
1
)
{\displaystyle (m+nr-1)\ldots (m+n(r-1)+1)}
=
2
m
+
n
r
−
1
Γ
(
n
)
π
∫
−
π
2
π
2
cos
m
+
n
r
−
1
θ
ε
(
m
+
n
(
r
−
2
)
+
1
)
i
θ
d
θ
=
2
m
+
n
r
−
1
π
∫
0
∞
∫
−
π
2
π
2
d
θ
d
z
cos
m
+
n
r
−
1
θ
ε
(
m
+
n
(
r
−
2
)
+
1
)
i
θ
.
z
n
−
1
ε
−
z
.
{\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
1
+
(
m
+
2
r
−
1
)
α
+
(
m
+
3
r
−
1
)
(
m
+
3
r
−
2
)
1.2
α
2
+
&
c
.
{\displaystyle 1+(m+2r-1)\alpha +{\frac {(m+3r-1)(m+3r-2)}{1.2}}\alpha ^{2}+\mathrm {\&c.} }
=
2
m
+
r
−
1
π
∫
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
)
;
{\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}}}