Integration of complex exponential function Today, my professor presented the following integral
\begin{align*}
    \int_{}^{} e^{ix}\mathrm{d}x = \frac{e^{ix}}{i}
.\end{align*}
where $i$ denotes the imaginary unit. I was a bit confused about this result, since we haven't introduced differentiation nor integration of complex valued functions. So my question is: How is the integration of complex functions defined?
 A: The other comments are not justified. But you can work this out with
$$\int e^{ix}dx=\int(\cos(x)+i\sin(x))dx=\int\cos(x)dx+i\int\sin(x)dx=\sin(x)-i\cos(x)+C=\dfrac{e^{ix}}i+C$$ where the integrals are real.
Update: an answer was modified and comments deleted since.
A: Here the $x$ in $dx$ is a real variable so it is real integration, the confusion also arises because the bounds are not explicitly marked, so the fact it is a real interval is somehow hidden.
Would it be $dz$ then this would be a completely different animal and you'll have to integrate on a contour and this would be complex integration.

It is not the nature of function that matters, but the nature of the variable.

So in this case the $i$ in the $e^{ix}$ just acts as a scalar, and it integrates as if it was $e^{\alpha x}$.
It we expand the integral we get:
$\begin{array}{lll}\displaystyle\int e^{ix}dx
&=\displaystyle\int(\cos(x)+i\sin(x))dx
&=\displaystyle\underbrace{\int\cos(x)dx}_\text{real integral}+\underbrace{i}_\text{scalar}\times\underbrace{\int\sin(x)dx}_\text{real integral}\\\\
&=\sin(x)+i(-\cos(x))
&=(i\sin(x)+\cos(x))\times\underbrace{(-i)}_{=\frac 1i}\\
&=\dfrac{e^{ix}}i\end{array}$
