Complex Riemann Integral is Contour Integral

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Theorem

Let $f: \R \to \C$ be a complex Riemann integrable function over some closed real interval $\left[{a \,.\,.\, b}\right]$.


Then:

$\displaystyle \int_a^b f \left({t}\right) \rd t = \int_\mathcal C f \left({t}\right) \rd t$

where:

the integral on the left hand side is a complex Riemann integral
the integral on the right hand side is a contour integral
$\mathcal C$ is a straight line segment along the real axis, connecting $a$ to $b$.


Proof

\(\displaystyle \int_a^b f \left({t}\right) \rd t\) \(=\) \(\displaystyle \int_a^b f \left({\theta \left({t}\right)}\right) \theta' \left({t}\right) \rd t\) $\quad$ Complex Integration by Substitution: $\theta \left({t}\right) = t$, $\theta' \left({t}\right) = 1$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \int_\mathcal C f \left({t}\right) \rd t\) $\quad$ Definition of Complex Contour Integral $\quad$

$\blacksquare$