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I'm reading Conway's complex analysis book and on page 62 he proved this theorem:

Theorem 1.9. If $\gamma$ is piecewise smooth and $f:[a,b]\to \mathbb C$ is continuous then $$\int_a^bfd\gamma=\int_a^bf(t)\gamma'(t)dt$$

However he made the following claim before this theorem:

The following theorem says that in this case we can find $\int fd\gamma$ by the methods of integration learned in calculus.

These functions are complex valued ones, I only know how to integrate the real valued ones.

Which calculus is the author mentioning?

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Closely related, though perhaps less thoroughly answered: Why do we need the fundamental theorem of calculus to solve this line integral? – hardmath Jan 10 at 13:37
up vote 6 down vote accepted

For $f:[a,b]\to\Bbb C$ continuous, we write $f(t)=u(t)+i\,v(t)$ for real-valued function $u,v:[a,b]\to\Bbb R$ and define $$\int_a^bf(t)dt:=\int_a^bu(t)dt+i\int_a^b v(t)dt.$$

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Yes, but why is this true? where can I see the proof of this claim formally? – user42912 Jan 10 at 13:10
@user42912 It is the definition of integral of functions $f:[a,b]\to\Bbb C$. Hence, does not require proof. For example, see Rudin Real and Complex Analysis. – Spenser Jan 10 at 13:12
@user42912 It goes through the same if you replace $\mathbb{C}$ by $\mathbb{R}^n$: the point is that to integrate $f : [a,b] \to \mathbb{R}^n$, you just integrate each component. – Ian Jan 10 at 13:15
This may be the definition of integration of a function $f:[a,b]\to \mathbb C$, but one might also have defined such integration directly by Riemann sums, for example. In that case one would have to prove that each Riemann sum splits into real and imaginary parts, and that going to the limit preserves this split. – Henning Makholm Jan 10 at 13:23
@HenningMakholm: Since Riemann integration theory depends on upper and lower sums for convergence, we should probably develop the theory on complex integration using real and imaginary components at the outset (since only $\mathbb{R}$ is ordered). Of course in hindsight continuity on a closed interval $[a,b]$ will make the results equal. – hardmath Jan 10 at 13:45

The right hand side is an ordinary integral on the real interval $[a,b]$, at least if the real and imaginary parts of the integrand are considered separately.

That is, the real part of the integrand and the imaginary part may by linearity of the integral operator be combined to give a complex definite integral that is otherwise in accord with a Riemann or a Lebesgue definition.

Note that both $f(t)$ and $\gamma'(t)$, the "slope" of the path of integration, will in general contribute to these real and imaginary parts:

$$ f(t) \gamma'(t) = u(t) + i v(t) $$

where $u,v:[a,b] \to \mathbb{R}$ are continuous.

This would be straightforward if both the real and imaginary parts of the parameterized path $\gamma:[a,b] \to \mathbb{C}$ were smooth (differentiable). In practice we often employ paths that are only piecewise-differentiable (but continuous), as when a "contour integral" is contrived with a keyhole or other convenient shape for avoiding singularities in the integrand $f$.

But piecewise-differentiable is good enough, since we then (typically without further discussion) break up the interval of integration $[a,b]$ into segments on which $\gamma$ is differentiable (in both real and imaginary components, as a function of $t$) and consider the definite integral as a sum of definite integrals on those segments that partition $[a,b]$.

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