and a corresponding formula for v in terms of V. If O be the centre of the circle, Q be the interior point z, P the point aE(iθ) of the circumference, and ω the angle which QP makes with OQ produced, this integral is at once found to be the same as
| u = | 1 | ∫ Udω − | 1 | ∫ Udθ |
| π | 2π |
of which the second part does not depend upon the position of z, and the equivalence of the integrals holds for every arc of integration.
Conversely, let U be any continuous real function on the circumference, U0 being the value of it at a point P0 of the circumference, and describe a small circle with centre at P0 cutting the given circle in A and B, so that for all points P of the arc AP0B we have |U − U0| < ε, where ε is a given small real quantity. Describe a further circle, centre P0 within the former, cutting the given circle in A′ and B′, and let Q be restricted to lie in the small space bounded by the arc A′P0B′ and this second circle; then for all positions of P upon the greater arc AB of the original circle QP² is greater than a definite finite quantity which is not zero, say QP² > D². Consider now the integral
| u′ = | 1 | ∫ U | (a² − r²) | dθ = | 1 | ∫ Udω − | 1 | ∫ Udθ, |
| 2π | a² + r² − 2ar cos (θ − φ) | π | 2π |
which we evaluate as the sum of two, respectively along the small arc AP0B and the greater arc AB. It is easy to verify that, for the whole circumference,
| U0 = | 1 | ∫ U0 | a² − r² | dθ = | 1 | ∫ U0 dω − | 1 | ∫ U0 dθ. |
| 2π | a² + r² − 2ar cos (θ − φ) | π | 2π |
Hence we can write
| u′ − U0 = | 1 | ∫ AP0B (U − U0)dω − | 1 | ∫ AP0B (U − U0)dθ + | 1 | ∫ AB (U − U0) | (a² − r²) | dθ. |
| 2π | 2π | 2π | QP² |
If the finite angle between QA and QB be called Φ and the finite angle AOB be called Θ, the sum of the first two components is numerically less than