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$\newcommand\la\lambda\newcommand\w{\mathfrak w}\newcommand\R{\mathbb R}$Numerical calculations and other considerations (The min of the mean of iid exponential variables) 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?

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  • $\begingroup$ Maybe it helps to set $s=(e^{itu}-1)/(it)$, identify the geometric series and thus write it as $i\int_{\mathbb R} \sum_{n=0}^\infty\left((e^{itu}-1)^{n+1}\,/\,(i t)^{n+2}\right)dt$. Having gotten rid of the fraction, I'd try to formally switch the integral and sum and see if you can compute the principal value. Maybe they are $\pi\,u^{n+1}$ and the sum gives you the result. $\endgroup$ – Nikolaj-K Nov 20 at 19:02
  • $\begingroup$ The numeric calculation with Mathematica NIntegrate[(1 - Exp[I*t*1/2])/(Exp[I*t*1/2] - 1 - I*1/2)/ t, {t, -1000, -0.001}, AccuracyGoal -> 3, PrecisionGoal -> 3] + NIntegrate[(1 - Exp[I*t*1/2])/(Exp[I*t*1/2] - 1 - I*1/2)/t, {t, 0.001, 1000}, AccuracyGoal -> 3, PrecisionGoal -> 3] produces $1.25389\, +2.51409 i$ and does not confirm your hypothesis. $\endgroup$ – user64494 Nov 20 at 19:07
  • $\begingroup$ Also NIntegrate[(1 - Exp[I*t*1/2])/(Exp[I*t*1/2] - 1 - I*1/2)/ t, {t, -10000, -0.0001}, AccuracyGoal -> 3, PrecisionGoal -> 3, WorkingPrecision -> 30] + NIntegrate[(1 - Exp[I*t*1/2])/(Exp[I*t*1/2] - 1 - I*1/2)/t, {t, 0.0001, 10000}, AccuracyGoal -> 3, PrecisionGoal -> 3, WorkingPrecision -> 30] produces $1.2564281632324901625528374684+2.51331913735615084972161764584 i$. $\endgroup$ – user64494 Nov 20 at 19:31
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    $\begingroup$ @user64494: It is no longer a hypothesis. See the two proofs below. $\endgroup$ – GH from MO Nov 20 at 19:36
  • $\begingroup$ @user64494 -- there is a way to avoid the need to take a principal value, which returns a value close to the expected answer (I worked this out in the answer box, it's a method which I have found quite useful). $\endgroup$ – Carlo Beenakker Nov 20 at 21:07
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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 $u\rightarrow 1$.


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.

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    $\begingroup$ Why are there no other poles? (I was thinking about the same argument and haven't been able so far to show this.) $\endgroup$ – Christian Remling Nov 20 at 17:53
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    $\begingroup$ Sorry, I still don't understand this. How does it follow that the solutions of $e^{-u\tau}+\tau =1$ are all real? $\endgroup$ – Christian Remling Nov 20 at 18:31
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    $\begingroup$ Thank you. I don't know why, before seeing your answer, I decided to deal with the poles in the lower half-plane. :-) $\endgroup$ – Iosif Pinelis Nov 20 at 19:18
  • $\begingroup$ The integration by parts in an improper integral should be grounded. In other case this is done in the L. Euler's style. $\endgroup$ – user64494 Nov 21 at 5:46
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$\newcommand\eps\varepsilon$ We want to show that, under $R\to\infty$ and $\eps\to 0+$, we have $$\int_{(-R,-\eps)\cup(\eps,R)} \frac{1-e^{itu}}{e^{itu}-1-it}\,\frac{dt}t=\pi i\,\frac u{1-u}+o(1).$$ Equivalently, $$\int_{(-R,-\eps)\cup(\eps,R)}\left(\frac{1-e^{itu}}{e^{itu}-1-it}+1\right)\,\frac{dt}t=\pi i\,\frac u{1-u}+o(1).$$ In other words, $$\int_{(-R,-\eps)\cup(\eps,R)}\frac{dt}{e^{itu}-1-it}=\pi\,\frac u{u-1}+o(1).$$ The integrand is holomorphic in an open set containing $\{t\in\mathbb{C}:\text{$\Im(t)\geq 0$ and $t\neq 0$}\}$, hence by Cauchy's theorem it suffices to show that $$\int_{\gamma(R)}\frac{dt}{e^{itu}-1-it}=-\pi+o(1)\qquad\text{and}\qquad \int_{\gamma(\eps)}\frac{dt}{e^{itu}-1-it}=\frac{\pi}{u-1}+o(1),$$ where $\gamma(r)$ is the semicircle in $\{t\in\mathbb{C}:\Im(t)\geq 0\}$ going from $r$ to $-r$. For large $r$, the integrand on $\gamma(r)$ is $i/t+O(1/t^2)$. For small $r$, the integrand on $\gamma(r)$ is $-i/(t(u-1))+O_u(1)$. The result follows.

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This is to detail Carlo Beenakker's assertion about the poles of the integrand. Suppose that $t=x+iy$ is such a pole, where $x$ and $y$ are real. Then $$1-y=e^{-uy}\cos ux,\quad x=e^{-uy}\sin ux.$$ Suppose that $y>0$. If $x=0$ then $1-y=e^{-uy}\ge1-uy$, so that $(u-1)y\ge0$, which contradicts the conditions $y>0$ and $u\in(0,1)$. So, $x\ne0$ and hence $$\frac{\sin ux}{ux}=\frac{e^{uy}}u>1,$$ which contradicts the inequality $\frac{\sin v}{v}\le1$ for all real $v\ne0$.

So, $y\le0$.

If now $y=0$ then $1=\cos ux$ and hence $x=\sin ux=0$.

Thus, the only pole $x+iy$ with $y\ge0$ is $0$.

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  • $\begingroup$ @GHfromMO : Yes indeed, this is much simpler. $\endgroup$ – Iosif Pinelis Nov 20 at 19:37
  • $\begingroup$ @ChristianRemling : Indeed. When I saw $t$ in the inequality $1\le|1+it|$ I somehow forgot that $t$ is complex. :-) So, it again looks like the assertion about the poles is not quite trivial. $\endgroup$ – Iosif Pinelis Nov 20 at 20:57
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    $\begingroup$ @IosifPinelis: For what it's worth, I tried it for about 10 minutes unsuccessfully (and then Carlos posted his answer), so I think it has to be something like the argument you give here. $\endgroup$ – Christian Remling Nov 20 at 21:52

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