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.

~ 115.

Illustration.

195

between infinite limits is no longer finite; hence we introduce the
hypothesis that m is at most equal to n - 1.
The second part only of each Pk has now to be considered in the
summation and all of them converge to zero for x = o.
In order to determine the value of Qk let us observe, that since a
was assumed < z, we have
c +27 1+ _i = < 2
for all values of == 0, 1,...,n - 1. For x = 0, Qk =tan- (tan,);
let us put it equal to, in whatever quadrant this be.
If, lie in the first quadrant (cos rz > 0, sin t > 0), Qk increases
as x does, it becomes + I- for x = cos,. For greater values of
x the argument is negative, for x = -oo we have Qk --.
If t be in the second quadrant (cos / < 0, sin u > 0), the argument is throughout negative, for x = o, Qk = Z.
If,t be in the third quadrant (cos, < 0, sin ~ < 0), the argument is throughout positive, for x -= o, Qk ==.
If, be in the fourth quadrant (cos, > 0, sin p < 0), Qk decreases as x increases, it is 3 Z for x = cos /z; it becomes less than
— rt for greater values of x and for x = oc, Qk == L. Accordingly:
Co k = n — 1
= n+ -^,,ck=- e__
_ __ __=__t a + 27~1+ 1x) a n2 /j
0
The summation is carried out as follows: ve have for any value of
< < 2r
e = e-n - 1 i Xk c e = neli te (i _t 1) - e' (e -_ 1 )
e '- e ee ' 1 --
el k t i _ k2 i. e 1 (ei..1)2
kk —O k=O
The second formula results by differentiating the first with respect to A.
Putting n 2 = 2mni we obtain:
k-9e =n- i 2 nz i i k= n -1 _2m n L ni
2keo e n _ 1 ik —o en u e n
2m rt i 2m i
k1 n 0 C 2 s in
Accordingly when in < n:
J x +dx- ^ 2ie- n 7 n e_ et, nr
2 e k c i 2 l e it C M n
k=0 sin.
Substituting for in this f and writingl a for the proper
rational fractioen _ —, we obtain the fundamen.tal definite integral:
+, lo 'Ula t f s I00rte
j d csin a (a-I)ai, _- v <K a < + -; see further ~ 159.
Jz + (-] eli' sin alt
13*