Last I had an exam and there was the following question: Find $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $\frac{1}{2\pi}\int\limits_{-\infty}^{\infty}(1+i)\mathrm{F}(\omega)e^{iwt}\mathrm{d}\omega = e^{-2t}H(t)$, where $\mathrm{F}(\omega)$ is the Fourier transform of $f$, and $H(t)$ is the Heaviside function.
The left side of the equation clearly has the form of an inverse transform, so I thought I'd just take the $(1+i)$ out from the integral, since it's a constant, so that $\frac{1}{2\pi}\int\limits_{-\infty}^{\infty}\mathrm{F}(\omega)e^{iwt}\mathrm{d}\omega = \frac{e^{-2t}H(t)}{1+i}$. Then $\mathrm{F}(\omega)(t)$ would be the transform of $\frac{e^{-2t}H(t)}{1+i}$ and therefore $f(t)$ would be $\frac{e^{-2t}H(t)}{1+i}$. But this function is not Real! (And we have not worked with Fourier transforms of functions of a complex variable)
Do you think there is a mistake in the question (it's not unusual at all, believe it or not!), or am I missing something or doing something wrong? Because if instead of $(1+i)$ there was something else, even a function of $\omega$, say $g(\omega)$, then I'd just transform $e^{-2t}H(t)$, divide it by $g(\omega)$, antitransform and get my $f(t)$, as long as that I don't get a complex function.