where [epsilon] is numerically less than unity. Now supposing [alpha] to diminish to zero we finally obtain _ / [oo] tan x [pi] | ----- dx = ---- _/ 0 x 2
(2) For another case, to illustrate a different point, we may take the integral _ / z^(a-1) | ------- dz, _/ 1 + z
wherein a is real quantity such that 0 < a < 1, and the contour consists of a small circle, z = rE(i[theta]), terminated at the points x = r cos [alpha], y = ± r sin [alpha], where [alpha] is small, of the two lines y = ± r sin [alpha] for r cos [alpha] =< x =< R cos ß, where R sin ß = r sin [alpha], and finally of a large circle z = RE(i[phi]), terminated at the points x = R cos ß, y = ± R sin ß. We suppose [alpha] and ß both zero, and that the phase of z is zero for r cos a =< x =< R cos ß, y = r sin [alpha] = R sin ß. Then on r cos [alpha] =< x =< R cos ß, y = -r sin [alpha], the phase of z will be 2[pi], and z^([alpha] - 1) will be equal to x^([alpha] - 1) exp (2[pi]i(a - 1)), where x is real and positive. The two straight portions of the contour will thus together give a contribution _ / R cos ß x^(a - 1) [1 - exp (2[pi]i[alpha])] | --------- dx. _/ r cos [alpha] 1 + x
It can easily be shown that if the limit of z[f](z) for z = 0 is zero, the integral [int] [f](z)dz taken round an arc, of given angle, of a small circle enclosing the origin is ultimately zero when the radius of the circle diminishes to zero, and if the limit of z[f](z) for z = [oo] is zero, the same integral taken round an arc, of given angle, of a large circle whose centre is the origin is ultimately zero when the radius of the circle increases indefinitely; in our case with [f](z) = z^([alpha] - 1)/(1 + z), we have z[f](z) = z^a/(1 + z), which, for 0 < a < 1, diminishes to zero both for z = 0 and for z = [oo]. Thus, finally the limit of the contour integral when r = 0, R = [oo] is _ / [oo] x^([alpha] - 1) [1 - exp (2[pi]i[alpha])] | --------------- dx. _/ 0 1 + x
Within the contour [f](z) is single valued, and has a pole at z = 1; at this point the phase of z is [pi] and z^(a - 1) is exp [i[pi](a - 1)] or - exp (i[pi]a); this is then the residue of [f](z) at z = -1; we thus have _ / [oo] x^(a - 1) [1 - exp (2[pi]ia)] | --------- dx = -2[pi]i exp (i[pi]a), _/ 0 1 + x