| | ∫ ½ π 0 F(z) | sin mz | dz − | π | F(0) | < ε. |
| sin z | 2 |
It has now been shown that when m is indefinitely increased ∫ π/2 0 F(z) (sin mz / sin z) dz − (π/2) F(0) has the limit zero.
Returning to the form (4), we now see that the limiting value of
| 1 | ∫ π/2 0 F(z) | sin mz | + | 1 | ∫ π/2 0 F(−z) | sin mz | dz is ½ {F(+0) + F(−0)}; |
| π | sin z | π | sin z |
hence the sum of n + 1 terms of the series
| 1 | ∫ l −l ƒ(x) dx + | 1 | Σ ∫ l −l ƒ(x¹) cos | nπ(x − x¹) | dx |
| 2l | l | l |
converges to the value ½ {ƒ(x + 0) + ƒ(x − 0)}, or to ƒ(x) at a point where ƒ(x) is continuous, provided ƒ(x) satisfies Dirichlet’s conditions for the interval from −l to l.
Proof that Fourier’s Series is in General Uniformly Convergent.—To prove that Fourier’s Series converges uniformly to its sum for all values of x, provided that the immediate neighbourhoods of the points of discontinuity of ƒ(x) are excluded, we have
| | ∫ π/2 F(z) | sin mz | dz − | π | F(0) | < πG(μ) + | 4 | {F(μ + 0) − F(0)} + | 4 | {F(½ π − 0) − F(0)} |
| sin z | 2 | m sin μ | m sin ξ¹ |
| < | πμ | {ƒ(x + 2μ) − ƒ(x)} + | 4 | {ƒ(x + 2μ) − ƒ(x)} + | 4 | {ƒ(x + π) − ƒ(x)}. |
| sin μ | m sin μ | m sin ξ¹ |