§ 1. Integrals of Holomorphic Functions
23
μ0
f (ζ)dζ =
μ1
f (ζ)dζ .
(3.8)
Suppose now that all paths μ s are closed, i.e. that μ s (1) = μ s (0) for all s. A
short calculation shows that the right hand side of (7) is still zero for all s.
Hence (8) follows again.
Without using these assumptions, the FT applied to relation (7) shows
that, by integrating over [0, 1],
μ1
f (ζ)dζ −
μ0
f (ζ)dζ =
1
0
f [μ s (1)] [μ 1 (1) − μ 0 (1)] ds −
(3.9)
−
1
0
f [μ s (0)] [μ 1 (0) − μ 0 (0)] ds .
μ
μ
μ
μ 0
1/3
2/3
1
μ
γ
γ
γ
γ
Fig. 3.2.
As μ 1 (1) − μ 0 (1) is the derivative of μ s (1) with respect to s, the first term is
just the integral of f along the path s → μ s (1), the second being the integral
of f along the path s → μ s (0). Therefore, the relation obtained means that,
the integral of f along the coherently oriented closed path γ drawn in figure
2 above is zero. This would be obvious if f had a primitive on G, which is,
however, not assumed. We will not generalize to any closed path : as a closed
path, γ is homotopic to the path consisting in traveling μ 0 twice in opposite
directions, and so is homotopic to a point. As for the fact that the integral
of f along γ is zero, it follows from Theorem 3 which will be proved shortly.
(iv) The homotopy invariance theorem. Paths μ 0 and μ 1 always being
admissible, suppose only that the homotopy σ deforming μ 0 into μ 1 is C
0 ; it
is no longer possible to differentiate integrals or even to write them. But a
given homotopy can be approached by linear homotopies and the preceding
point can be used, which, as will be seen, again leads to the same results.
23
μ0
f (ζ)dζ =
μ1
f (ζ)dζ .
(3.8)
Suppose now that all paths μ s are closed, i.e. that μ s (1) = μ s (0) for all s. A
short calculation shows that the right hand side of (7) is still zero for all s.
Hence (8) follows again.
Without using these assumptions, the FT applied to relation (7) shows
that, by integrating over [0, 1],
μ1
f (ζ)dζ −
μ0
f (ζ)dζ =
1
0
f [μ s (1)] [μ 1 (1) − μ 0 (1)] ds −
(3.9)
−
1
0
f [μ s (0)] [μ 1 (0) − μ 0 (0)] ds .
μ
μ
μ
μ 0
1/3
2/3
1
μ
γ
γ
γ
γ
Fig. 3.2.
As μ 1 (1) − μ 0 (1) is the derivative of μ s (1) with respect to s, the first term is
just the integral of f along the path s → μ s (1), the second being the integral
of f along the path s → μ s (0). Therefore, the relation obtained means that,
the integral of f along the coherently oriented closed path γ drawn in figure
2 above is zero. This would be obvious if f had a primitive on G, which is,
however, not assumed. We will not generalize to any closed path : as a closed
path, γ is homotopic to the path consisting in traveling μ 0 twice in opposite
directions, and so is homotopic to a point. As for the fact that the integral
of f along γ is zero, it follows from Theorem 3 which will be proved shortly.
(iv) The homotopy invariance theorem. Paths μ 0 and μ 1 always being
admissible, suppose only that the homotopy σ deforming μ 0 into μ 1 is C
0 ; it
is no longer possible to differentiate integrals or even to write them. But a
given homotopy can be approached by linear homotopies and the preceding
point can be used, which, as will be seen, again leads to the same results.