$\newcommand\la\lambda\newcommand\w{\mathfrak w}\newcommand\R{\mathbb R}$Numerical calculations and other considerations ([The min of the mean of iid exponential variables](https://mathoverflow.net/questions/376819/the-min-of-the-mean-of-iid-exponential-variables/377007#377007)) suggest that  

$$\int\_\R \frac{1-e^{itu}}{e^{itu}-1-it}\,\frac{dt}t=\pi i\,\frac u{1-u}$$
for $u\in(0,1)$, with the integral understood in the principal value sense. However, I have not been able to prove this, even with the help of Mathematica.


How can this be proved?



An integral identity
I would close the contour in the upper half of the complex plane, the principal value picks up $i\pi$ times the residue$^\ast$ at $t=0$, which is $u/(1-u)$. There are no other poles.$^{\ast\ast}$


$^\ast$ $\frac{1-e^{i t u}}{e^{i t u}-i t-1}=\frac{u}{1-u}+{\cal O}(t^2).$


$^{\ast\ast}$ poles are at $t=i\tau$ with $e^{-\tau u}+\tau=1$ (excluding $\tau=0$, which is canceled by the numerator); these remain at $\tau<0$ for all $u\in(0,1)$, approaching $-2(1-u)$ for $u\rightarrow 1$.




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In the comments there was an issue with the numerical evaluation. Principal value integrals of this type can be evaluated more accurately by replacing $1/t$ by $\frac{d\log |t|}{dt}$ and carrying out a partial integration. This gives
$$\int\_{-\infty}^\infty dt\,\frac{1-e^{itu}}{e^{itu}-1-it}\,\frac{1}t=
-2i\Im\int\_{0}^\infty dt\,\ln|t|\frac{d}{dt}\frac{1-e^{itu}}{e^{itu}-1-it}.$$
For the case $u=1/2$ considered in the comments, Mathematica gives 3.1406.


