| 2 ∫ ∞0 | tan x | dx − (π − 2α) − 2αε = 0, |
| x |
where ε is numerically less than unity. Now supposing α to diminish to zero we finally obtain
| ∫ ∞0 | tan x | dx = | π | . |
| x | 2 |
(2) For another case, to illustrate a different point, we may take the integral
| ∫ | za−1 | dz, |
| 1 + z |
wherein a is real quantity such that 0 < a < 1, and the contour consists of a small circle, z = rE(iθ), terminated at the points x = r cos α, y = ± r sin α, where α is small, of the two lines y = ± r sin α for r cos α ⋜ x ⋜ R cos β, where R sin β = r sin α, and finally of a large circle z = RE(iφ), terminated at the points x = R cos β, y = ±R sin β. We suppose α and β both zero, and that the phase of z is zero for r cos a ⋜ x ⋜ R cos β, y = r sin α = R sin β. Then on r cos α ⋜ x ⋜ R cos β, y = −r sin α, the phase of z will be 2π, and zα − 1 will be equal to xα − 1 exp [2πi(a − 1)], where x is real and positive. The two straight portions of the contour will thus together give a contribution
| [1 − exp(2πiα)] ∫ R cos βr cos α | xa−1 | dx. |
| 1 + x |
It can easily be shown that if the limit of zƒ(z) for z = 0 is zero, the integral ∫ƒ(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 zƒ(z) for z = ∞ 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 ƒ(z) = zα−1/(1 + z), we have zƒ(z) = za/(1 + z), which, for 0 < a < 1, diminishes to zero both for z = 0 and for z = ∞. Thus, finally the limit of the contour integral when r = 0, R = ∞ is
| [1 − exp(2πiα)] ∫ ∞0 | xα−1 | dx. |
| 1 + x |
Within the contour ƒ(z) is single valued, and has a pole at z = 1; at this point the phase of z is π and za−1 is exp [iπ(a − 1)] or − exp(iπa); this is then the residue of ƒ(z) at z = −1; we thus have