| 1 | ∫ π/2 0 F(z) | sin mz | dz + | 1 | ∫ π/2 0 F(−z) | sin mz | dz. |
| π | sin z | π | sin z |
(4)
We require therefore to find the limiting value, when m is indefinitely increased, of ∫ π/2 0 F(z) (sin mz) / (sin z) dz; the form of the second integral being essentially the same. This integral, or rather the slightly more general one ∫ h 0 F(z) (sin mz) / (sin z) dz, when 0 < h ≦ ½π, is known as Dirichlet’s integral. If we write X(z) = F(z) (z / sin z), the integral becomes ∫ h 0 X(z) (sin mz) / z dz, which is the form in which the integral is frequently considered.
The Second Mean-Value Theorem.—The limiting value of Dirichlet’s integral may be conveniently investigated by means of a theorem in the integral calculus known as the second mean-value theorem. Let a, b be two fixed finite numbers such that a < b, and suppose ƒ(x), φ(x) are two functions which have finite and determinate values everywhere in the interval except for a finite number of points; suppose further that the functions ƒ(x), φ(x) are integrable throughout the interval, and that as x increases from a to b the function ƒ(x) is monotone, i.e. either never diminishes or never increases; the theorem is that
∫ b a ƒ(x) φ(x) dx = ƒ(a + 0) ∫ ξ a φ(x) dx + ƒ(b − 0) ∫ b ξ φ(x) dx
when ξ is some point between a and b, and ƒ(a), ƒ(b) may be written for ƒ(a + 0), ƒ(b − 0) unless a or b is a point of discontinuity of the function ƒ(x).
To prove this theorem, we observe that, since the product of two integrable functions is an integrable function, ∫ b a ƒ(x) φ(x) dx exists, and may be regarded as the limit of the sum of a series
ƒ(x0) φ(x0) (x1 − x0) + ƒ(x1) φ(x1) (x2 − x1) + ... + ƒ(xn−1) φ(xn−1) (xn − xn−1)
where x0 = a, xn = b and x1, x2 ... xn−1 are n − 1 intermediate points. We can express φ(xr) (xr+1 − xr) in the form Yr+1 − Yr, by putting
Yr = ΣK=r K=1 φ (xK-1) (xK − xK−1), Y0 = 0.