An introduction to the study of the elements of the differential and integral calculus. From the German of the late Axel Harnack, With the permission of the author.

292 Examples on the calculation of definite integrals. Bk. III. ch. VII.
also the product F(a), converges absolutely for all finite values of a
except negative integers.
165. Legendre's series for calculating IF(a).
Writing Equation VIII) in the form:
dF(1 -+- a) 1 1 1
' 2 F ( e +a + 2 + (a >-1)
da2 a + 1)' (a -+ 2)2 (a + 3)2
and differentiating it n - 2 times, which is allowable since the derived
series likewise converge absolutely, the result is:
1 dlr(l - +a) _ (- )n 11 1
I n da a 12 (a + 2') ( a+ 3)
i I 1 1
Let the sum - + -t- - + - + 7. (~ 47 foot-note p. 82), be
ln +2n gn 4n
denoted by S,, we have:
1 fdn lr( +a) S_ 1 S)n
Ln ' da' n
a=O
further:
{dlr(l +a)} _ r'(l) = and: Ir(l)= 0.
a=0
Therefore by Mac Laurin's theorem:
a2 as a4 r L 1 c., ^ [ a)
IF(l +a)=aC + - S2 -- 3- S3 - + 4 a dn F(,.- a ).
2l f )a- 2 3 83- 4 in dan
Oa
The remainder converges to zero when the absolute amount of a is
less than 1, accordingly, omitting the remainder, the infinite series is
absolutely convergent for all values of a between - 1 and + 1.
But this series is unsuitable for numerical calculation, because the
coefficients S do not decrease rapidly enough, and moreover the value
of C is still unknown. A more rapidly convergent series is found
by expanding the value of 1(1 + a) and adding:
a2 a3 a,
==- (1 a)+ a-+- 3 +.
3 4 +
thus:
1r(l + a) =- I(] + a)+ a+( ( S + - - 2 /Se — S - 1)C3
+ (S-1)a --
likewise, taking a with the opposite sign:
/F(1 - a) ( - - a ) - a(1 + C) + - (- 1)a+ (3 - )a
(S4 _ 1)a4+..
Now because by Equation V):
Ir(l + a) + Ir(l - a)=-Z I a,
sin Tt a