Inverse Fourier Transform gives a complex function when it shouldn't 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. 
 A: First of all, I think the title is a misleading. The Fourier transform (and its inverse) is defined as an operator on complex functions of real variable (i.e. $\mathscr{F}:\mathbb{C^R\rightarrow C^R}$, $f\mapsto F$, where $f, F: \mathbb{R\rightarrow C}$ are integrable functions). For this reason I think (contrary to what the title suggests) that the inverse Fourier transform should yield a complex function, it just so happens that this result can sometimes lie in the subset of real functions. This point has been made clear in the comments.
Now, you've calculated yourself the function $f$ that satisfies the expression which is in can be recast as the inverse Fourier transform of $F$. Note $f$ has a transform:$$F(\omega)=\frac{1}{2+i\omega}\cdot\frac{1}{1+i}=\frac{1}{(2-\omega)+i(2+\omega)}$$As you can check. 
You can also see in this other MSE post that for most "reasonable" functions the direct and inverse Fourier transforms are unique.Both $f$ and $F$ are piecewise continuous functions which do satisfy such uniqueness property. This means that the function $f$ you got (from the inverse transform expression) is the only possible answer.
So, I do think there is a mistake in the question as the result is a complex function rather than a real one.
