I want to calculate the integral $$I_n(\omega)=\int_{-\infty}^{\infty} \log^n\left(it-\omega\right) \, {\rm d}t$$ by analytic regularization where $\omega >0$ and $n \in \mathbb{N}$. One possibility is to calculate $$J_\omega(\epsilon)=\int_{-\infty}^{\infty} \left(it-\omega\right)^\epsilon \, {\rm d}t$$ for $\epsilon<-1$ and then $$I_n(\omega)=J_{\omega}^{(n)}(0) = \frac{{\rm d}^n}{{\rm d}\epsilon^n} \, \frac{\omega^{1-\epsilon}\,2\sin(\epsilon \pi)}{\epsilon-1}\Bigg|_{\epsilon=0}$$ where the superscript refers to the $n$-th derivative. Another possibility is to introduce a Cut-Off $T$ in the original integral and calculate the proper integral $$I_{n}^{T}(\omega)=\int_{-T}^{T} \log^n\left(it-\omega\right) \, {\rm d}t \, .$$ Finally expand the result in an asymptotic series in $T$ and remove the divergent part; then take the limit $T\rightarrow \infty$.
Now with the integral at hand it seems that the latter version gives the correct result, but the first one namely $J_\omega^{(n)}(0)$ only coincides with the regularization by cut-off if $n$ is even. When $n$ is odd the results differ by $\frac{\omega \pi^n}{2^{n-1}}$.
What's the reason for this???