Making a Fourier Transform converge

Question

Consider for example the following function \begin{equation} f(x)=(e^x+1)^{ik_1}\,. \end{equation} For $k_1 \in \mathbb{R}$, this is clearly not absolutely integrable and thus its Fourier transform should be a distribution. Is it possible to write that distribution explicitly? Besides that, if we write the Fourier transform explicitly \begin{equation} \mathcal{F}(f(x))=\int_{-\infty}^{+\infty}dx (e^x+1)^{ik_1}e^{ix k_2}\,, \end{equation} we can see we can make it converge to a standard function by setting $k_1+k_2 \to k_1+k_2+i\epsilon$ and $k_2 \to k_2 -i\epsilon$ with $\epsilon>0$. We can then do the Fourier transform and send $\epsilon \to 0$. Such $i\epsilon$ prescriptions are typical in physics and I was trying to understand if I should think of the object defined in this way as an entirely different object from $\mathcal{F}(f(x))$ which is a distribution or if there is usually a general relationship between the two. I cannot find out if such a relationship exists because I do not know how to write $\mathcal{F}(f(x))$ down without adopting such prescription that makes it converge to a standard function. Naively, I would be tempted to write \begin{equation} \mathcal{F}(f(x))=\delta(k_1+k_2)\,. \end{equation} But this is only strictly correct if I had $f(x)=e^{ik_1x}$ in which case I would not have an $i\epsilon$ prescription ensuring convergence. So I wonder what the factor of $1$ is actually doing to the distribution. Thanks.