2n + 1
sin ------ (x' - x)
2
= -------------------.
2 sin 1/2(x' - x)
Hence, with the new notation, the sum of the first n+1 terms of (3) is
_
1 / [pi] sin (2n + 1)/2 (x' - x)
---- | [f](x') ----------------------- dx'.
[pi]_/-[pi] 2 sin 1/2 (x' - x)
If we suppose [f](x) to be continued beyond the interval -[pi] to
[pi], in such a way that [f](x) = [f](x + 2[pi]), we may replace the
limits in this integral by x + [pi], x-[pi] respectively; if we then
put x' - x = 2z, and let [f](x') = [F](z), the expression becomes
1/[pi] [int][-[pi]/2 to [pi]/2] F(z) (sin mz/sin z) dz, where m = 2n +
1; this expression may be written in the form
_ _
1 /[pi]/2 sin mz 1 /[pi]/2 sin mz
---- | F(z) ------ dz + ---- | F(-z) ------ dz. (4)
[pi]_/ 0 sin z [pi]_/ 0 sin z
We require therefore to find the limiting value, when m is
indefinitely increased, of [int][0 to [pi]/2] 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 [int][0 to h]
F(z)(sin mz/sin z) dz, when 0 < h <= 1/2[pi], is known as Dirichlet's
integral. If we write X(z)= F(z)(z/sin z), the integral becomes
[int][0 to h] 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 [f](x), [phi](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 [f](x), [phi](x) are integrable throughout the
interval, and that as x increases from a to b the function [f](x) is
monotone, i.e. either never diminishes or never increases; the theorem
is that
_ _ _
/ b /[xi] / b
| [f](x) [phi](x) dx = [f](a + 0) | [phi](x) dx + [f](b - 0) | [phi](x) dx
_/ a _/ a _/[xi]
when [xi] is some point between a and b, and [f](a), [f](b) may be
written for [f](a + 0), [f](b - 0) unless a or b is a point of
discontinuity of the function [f](x).
To prove this theorem, we observe that, since the product of two
integrable functions is an integrable function, [int][a to b] [f](x)
[phi](x) dx exists, and may be regarded as the limit of the sum of a
series [f](x0) [phi](x0) (x1 - x0) + [f](x1) [phi](x1) (x2 - x1) + ...
+ [f]x(n-1) [phi]x_(n-1) (x_n - x_(n-1)) where x0 = a, x_n = b and x1,
x2 ... x_(n-1) are n - 1 intermediate points. We can express
[phi](x_r) (x_(r+1) - x_r) in the form Y_(r+1) - Y_r, by putting
K=r
Y_r = [Sigma] [phi](x_(K-1)) (x_K - x_(K-1)), Y0 = 0.
K=1