In the limit T/R → ∞ , when µ → 0, this equation simplifies to When z = s runs along the real axis, the variable ω runs along the boundary of the rectangle with s = − 1 / √ µ, − √ µ, + √ µ, +1 / √ µ ( µ < 1) mapped, respectively, onto the vertices of the rectangle: ( − R/ 2 , T/ 2) , ( − R/ 2 , − T/ 2) , ( R/ 2 , − T/ 2) , ( R/ 2 , T/ 2). The given choice of the argument of the mapping preserves the symmetry s → 1 /s . When z has positive imaginary part, the coordinates take their values inside the rectangle. These coordinates are conformal. For this reason we have whose implementation for the function t ∗ ( s ) is discussed below. The boundary contour given by Eq. (23) satisfies Douglas’ minimization (see Appendix B, Eq. (B3)). Correspondingly, the Douglas integral as it should. We have verified these two equations numerically. A natural parametrization of the boundary of a rectangle is through τ ∈ S $^{1}$: and analogously for negative τ . Introducing we rewrite Eq. (25) as 6 t = tan τ 2 , (26) x$_{1}$ = T t 1 − t 2 , x$_{2}$ = − T 2 − √$_{T}$ 2 + R 2 − T R ≤ t ≤ √$_{T}$ 2 + R 2 − T R (27a) x$_{1}$ = R 2 , x$_{2}$ = R t 2 − 1 4 t √$_{T}$ 2 + R 2 − T R ≤ t ≤ √$_{T}$ 2 + R 2 + T R (27b) x$_{1}$ = T t t 2 − 1 , x$_{2}$ = T 2 √$_{T}$ 2 + R 2 + T R ≤ t < + ∞ . (27c) x$_{1}$ = T 2 tan τ , x$_{2}$ = − T 2 − arctan R T ≤ τ ≤ arctan R T (25a) x$_{1}$ = R 2 , x$_{2}$ = − R 2 cot τ arctan R T ≤ τ ≤ π − arctan R T (25b) x$_{1}$ = T 2 tan τ , x$_{2}$ = T 2 π − arctan R T ≤ τ < π (25c) 1 4 π ∫ $^{+}$∞ −∞ d s ∫ $^{+}$∞ −∞ d s ′ [ x ( t ∗ ( s$_{1}$ )) − x ( t ∗ ( s$_{2}$ ))] 2 ( s s ′ ) 2 = 2 A $^{2}$K ( µ ) K (√ 1 − µ 2 ) = RT (24) x$_{1}$ ( t ∗ ( s )) = A Re F ( s √ µ , µ ) , x$_{2}$ ( t ∗ ( s )) = A Im F ( s √ µ , µ ) − AK ( √ 1 − µ 2 ) 2 , (23) X$_{1}$ ( z ) = A Re F ( z √ µ , µ ) , X$_{2}$ ( z ) = A Im F ( z √ µ , µ ) − AK ( √ 1 − µ 2 ) 2 (22) π T R = ln 4 µ . (21)