Complex Riemann Integral is Contour Integral

Theorem

Let $f: \R \to \C$ be a complex Riemann integrable function over some closed real interval $\closedint a b$.

Then:

$\ds \int_a^b \map f t \rd t = \int_\CC \map f t \rd t$

where:

the integral on the left hand side is a complex Riemann integral

the integral on the right hand side is a contour integral

$\CC$ is a straight line segment along the real axis, connecting $a$ to $b$.

Proof

\(\ds \int_a^b \map f t \rd t\)

\(=\)

\(\ds \int_a^b \map f {\map \theta t} \map {\theta'} t \rd t\)

Complex Integration by Substitution: $\map \theta t = t$, $\map {\theta'} t = 1$

\(\ds \)

\(=\)

\(\ds \int_\CC \map f t \rd t\)

Definition of Complex Contour Integral

$\blacksquare$