254 Mr. Babbage’s new methods of investigating the 
=• A 
\n + A 
+ A rtn , flV + &C ' 
^ (sin 6)" 1 (sin 20)” ~T “ (sin 30) n 
(0 
If after the integration we had put v instead of v, and then 
v= cos 9 + V — i sin 9 , we should have found 
”= A (TTef + A (sin 2 ep + A (Fin 3^ + &C * 
123 
—i —1 —3 
A (sin 0)*" ~l“ A (sin 20/ 1 ”i“ A (sin 30)” ( 2 ) 
Neither of the integrals exhibited in ( 1 ) and ( 2) are inte- 
grate in the most simple cases, and it is only by their com- 
bination that I have been able to obtain the sums of any series. 
Let us suppose A = i, A = — 1, A = i, A = — 1, &c. then 
2 3 4 
ZJC -ftt 
-2 r; 
V 
-Kx 
V 
V 
2 S + tt 
and 4” “ ''= : also let » = 1 ; 
I+T/ T l+V 
then the difference of the two series is 
{ 22+ J — 2Z — I "I "* „z 3 “ 3 
I+x/ 2^+i \+ v — 1 J sin0 sin 20 sin 38 
but the integral on the left side of this equation becomes 
(21/— 1) S ( 1 ) which is equal to — l ( z-\-b ) : hence, since 
- 2 z—n 
Zl 
= we have 
2 log V 
-* -3 
X 1 — X 1 X 5 X , o 
+ T7T7. + &0. 
sin 20 1 sin 3© 
.If# = 1, b= 0, and since log v — 9 \/ — 1, our series becomes 
— See. 
_I _2 .3 
log a; x — x x*—x ^ x s —x 
(3) 
0 sin 0 sin 20 * sin 30 
let x = cos (5'jh y/— 1, sin (f ; then, since log x will become 
9 ' \/ — 1, by dividing both sides by 2 s/ — 1, we shall have 
0' sin 6' sin 20' , sin 20' 0 , \ 
sin 
sin 20 
sin 30 
&C. 
The series (3) is integrable when multiplied by and this 
operation may be repeated any number of times; the first 
