where α < π, and < 2/π where α ≧ π. It follows that
| | ∫ β α | sinθ | dθ | ≦ π, provided 0 ≦ α < β. |
| θ |
To find the limit of ∫ π/2 0 F(z) (sin mz / sin z) dz, we observe that it may be written in the form
| F(0) ∫ π/2 0 | sin mz | dz + ∫ μ 0 {F(z) − F(0)} | sin mz | dz + ∫ π/2 μ {F(z) − F(0)} | sin mz | dz |
| sin z | sin z | sin z |
where μ is a fixed number as small as we please; hence if we use lemma (1), and apply the second mean-value theorem,
| ∫ π/2 0 F(z) | sin mz | dz − | π | F(0) = ∫ μ 0 {F(z) − F(0)} | z | sin mz | dz | |
| sin z | 2 | sin z | z |
| + {F(μ + 0) − F(0)} ∫ ξ1 μ | sin mz | dz + [F (½ π − 0) − F(0)] ∫ π/2 ξ1 | sin mz | dz |
| sin z | sin z |
when ξ¹ lies between μ and ½ π. When m is indefinitely increased, the two last integrals have the limit zero in virtue of lemma (2). To evaluate the first integral on the right-hand side, let G(z) = {F(z) − F(0)} (z / sin z), and observe that G(z) increases as z increases from 0 to μ, hence if we apply the mean value theorem
| | ∫ μ 0 G(μ) | sin mz | dz | = | G(μ) ∫ μ ξ | sin mz | dz | = | G(μ) ∫ mμ mξ | sinθ | dθ | < πG(μ), |
| z | z | θ |
where 0 < ξ < μ, since G(z) has the limit zero when z = 0. If ε be an arbitrarily chosen positive number, a fixed value of μ may be so chosen that πG(μ) < ½ ε, and thus that | ∫ μ 0 G(z) (sin mx / z) dz | < ½ ε. When μ has been so fixed, m may now be so chosen that