| 1 | l − | 2l | ( cos | 2πx | + | 1 | cos | 6πx | + | 1 | cos | 10πx | + ... ) |
| 4 | π² | l | 3² | l | 5² | l |
This series represents for general values of x the ordinate of the set of broken lines in fig. 4.
| Fig. 4. |
Dirichlet’s Integral.—The method indicated by Fourier, but first carried out rigorously by Dirichlet, of proving that, with certain restrictions as to the nature of the function ƒ(x), that function is in general represented by the series (3), consists in finding the sum of n+1 terms of that series, and then investigating the limiting value of the sum, when n is increased indefinitely. It thus appears that the series is convergent, and that the value towards which its sum converges is ½ {ƒ(x + 0) + ƒ(x − 0)}, which is in general equal to ƒ(x). It will be convenient throughout to take −π to π as the given interval; any interval −l to l may be reduced to this by changing x into lx / π, and thus there is no loss of generality.
We find by an elementary process that
½ + cos (x − x′) + cos 2(x − x′) + ... + cos n(x − x′)
| = | sin ½ (2n + 1) (x′ − x) | . |
| 2 sin ½ (x′ − x) |
Hence, with the new notation, the sum of the first n+1 terms of (3) is
| = ∫ π −π ƒ(x′) | sin ½ (2n + 1) (x′ − x) | dx′. |
| 2 sin ½ (x′ − x) |
If we suppose ƒ(x) to be continued beyond the interval −π to π, in such a way that ƒ(x) = ƒ(x + 2π), we may replace the limits in this integral by x + π, x − π respectively; if we then put x′ − x = 2z, and let ƒ(x′) = F(z), the expression becomes 1/π ∫ π/2 −π/2 F(z) (sin mz) / (sin z) dz, where m = 2n + 1; this expression may be written in the form