∫ β α sin mzdz = 1∫ γ α sin mz dz + 1∫ β γ sin mz dz
sin z sin α sin β

where α < γ < β, hence

| ∫ β α sin mzdz | < 2( 1+ 1) < 4;
sin z msin α sin β m sin α

a precisely similar proof shows that | ∫ β α (sin mz / z) dz | < 4 / mα, hence the integrals ∫ β α (sin mz / sin z) dz, ∫ β α (sin mz / z) dz, converge to the limit zero, as m is indefinitely increased.

3. If α > 0, | ∫ ∞ α (sin θ / θ) dθ | cannot exceed ½ π. For by the mean-value theorem | ∫ h α (sin θ / θ) dθ | < 2/α + 2/h,

hence | Lh = ∞ ∫ h α (sin θ / θ) dθ | ≦ 2/α

in particular if α ≧ π | ∫ ∞ α (sin θ / θ) dθ | ≦ 2/π < π/2.

Again d/dα ∫ ∞ α (sin θ / θ) dθ = − (sin α) / α, α > 0,

therefore ∫ ∞ α (sin θ / θ) dθ increases as α diminishes, when θ < α < π;

but lim α=0∫ ∞ α (sin θ / θ) dθ = π/2, hence | ∫ ∞ α (sin θ / θ) dθ | < π/2,