# An integral identity

$$\newcommand\la\lambda\newcommand\w{\mathfrak w}\newcommand\R{\mathbb R}$$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?

• 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. – Nikolaj-K Nov 20 at 19:02
• 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. – user64494 Nov 20 at 19:07
• 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 $1.2564281632324901625528374684+2.51331913735615084972161764584 i$. – user64494 Nov 20 at 19:31
• @user64494: It is no longer a hypothesis. See the two proofs below. – GH from MO Nov 20 at 19:36
• @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). – Carlo Beenakker Nov 20 at 21:07

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.

• Why are there no other poles? (I was thinking about the same argument and haven't been able so far to show this.) – Christian Remling Nov 20 at 17:53
• Sorry, I still don't understand this. How does it follow that the solutions of $e^{-u\tau}+\tau =1$ are all real? – Christian Remling Nov 20 at 18:31
• Thank you. I don't know why, before seeing your answer, I decided to deal with the poles in the lower half-plane. :-) – Iosif Pinelis Nov 20 at 19:18
• The integration by parts in an improper integral should be grounded. In other case this is done in the L. Euler's style. – user64494 Nov 21 at 5:46

$$\newcommand\eps\varepsilon$$ 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.

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 $$0$$.

• @GHfromMO : Yes indeed, this is much simpler. – Iosif Pinelis Nov 20 at 19:37
• @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. – Iosif Pinelis Nov 20 at 20:57
• @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. – Christian Remling Nov 20 at 21:52