| ∫ 1/2π−ε0 | tan x | dx, ∫ 3/2π−ζ1/2π+ε | tan x | dx, ∫ 5/2π−σ3/2π+ζ | tan x | dx, ... |
| x | x | x |
Now draw a contour consisting in part of the whole of the positive and negative real axis from x = −nπ to x = +nπ, where n is a positive integer, broken by semicircles of small radius whose centres are the points x = ±½π, x = ±¾π, ... , the contour containing also the lines x = nπ and x = −nπ for values of y between 0 and nπ tan α, where α is a small fixed angle, the contour being completed by the portion of a semicircle of radius nπ sec α which lies in the upper half of the plane and is terminated at the points x = ±nπ, y = nπ tan α. Round this contour the integral ∫ [(tan z / z)] dz has the value zero. The contributions to this contour integral arising from the semicircles of centres −½(2s − 1)π, + ½(2s − 1)π, supposed of the same radius, are at once seen to have a sum which ultimately vanishes when the radius of the semicircles diminishes to zero. The part of the contour lying on the real axis gives what is meant by 2 ∫ nπ0 [(tan x / x)] dx. The contribution to the contour integral from the two straight portions at x = ±nπ is
| ∫ nπ tan α0 idy ( | tan iy | − | tan iy | ) |
| nπ + iy | −nπ + iy |
where i tan iy, = −[exp(y) − exp(−y)]/[exp(y) + exp(−y)], is a real quantity which is numerically less than unity, so that the contribution in question is numerically less than
| ∫ nπ tan α0 dy | 2nπ | , that is than 2α. |
| n²π² + y² |
Finally, for the remaining part of the contour, for which, with R = nπ sec α, we have z = R(cos θ + i sin θ) = RE(iθ), we have
| dz | = idθ, i tan z = | exp(−R sin θ) E(iR cos θ) − exp(R sin θ) E(−iR cos θ) | ; |
| z | exp(−R sin θ) E(iR cos θ) + exp(R sin θ) E(−iR cos θ) |
when n and therefore R is very large, the limit of this contribution to the contour integral is thus
− ∫ π−αα dθ = − (π − 2α).
Making n very large the result obtained for the whole contour is