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Show that $\int_{-\infty}^{0^+}\frac{\mathrm{Log}(t)}{e^{-t}-1}dt=0$ where the integral is over a contour of the Hankel-type. What I mean is that the contour looks like this but reflected across the imaginary axis so that it comes from $-\infty$ below the real axis, loops around the origin in a counter clockwise way with radius $\varepsilon$, and then continues to $-\infty$ above the real axis. contour

I am meant to take the limit as $\varepsilon\rightarrow0$. What I have so far is to break the integral over the entire contour up into three integrals, one for each ray to $-\infty$ and one for the circular section near the origin:

$\int\limits_{-\infty}^{(0^+)}\frac{\mbox{Log}(t)}{e^{-t}-1}dt=\int\limits_{-\infty}^{-\varepsilon}\frac{\mbox{Log}(t)}{e^{-t}-1}dt+\int\limits_{-\varepsilon}^{-\infty}\frac{\mbox{Log}(t)}{e^{-t}-1}dt+\int_{\left|t\right|=\varepsilon}\frac{\mbox{Log}(t)}{e^{-t}-1}dt$

Now I know that for the first integral $t=\tau e^{i\pi}$ and for the second integral $t=\tau e^{-i\pi}$ with $0<\tau<\infty$. Substituting this gives $\int\limits_{-\infty}^{-\varepsilon}\frac{\mbox{Log}(t)}{e^{-t}-1}dt+\int\limits_{-\varepsilon}^{-\infty}\frac{\mbox{Log}(t)}{e^{-t}-1}dt=\int_{\infty}^{\varepsilon}\left(\ln|t|+i\pi-\ln|t|-i(-\pi)\right)(e^{\tau}-1)^{-1}(-d\tau)$

Not really sure where to go from here.

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  • $\begingroup$ note that for any $Re(s) > 1$ : $\displaystyle\int_0^{+\infty} \frac{t^{s-1}}{e^t -1} dt = \int_{-\infty}^{+\infty} \frac{e^{su}}{e^{e^u}-1} du$ and hence with $C$ your complicated contour : $\displaystyle\int_C \frac{t^{s-1}}{e^t -1} dt = (1-e^{2 i \pi s}) \int_0^{+\infty} \frac{t^{s-1}}{e^t -1} dt = \int_{-\infty}^{+\infty} + \int_{+\infty}^{2i \pi + \infty}+ \int_{2i \pi + \infty}^{2 i \pi - \infty}+\int_{2i \pi - \infty}^{- \infty} \frac{e^{su}}{e^{e^u}-1} du$ i.e. after a change of variable $t = e^u$, the complicated contour $C$ becomes a simple rectangle without any argument problem $\endgroup$ – reuns Apr 16 '16 at 7:25
  • $\begingroup$ Right which is expected since by Cauchy's theorem one can deform the integral path as long it doesn't run into any singularities or branch cuts, which the rectangular path satisfies. Nonetheless, I am still unsure how to prove the result for the rectangular path. I think the Hankel type path was chosen because it might be easier than the rectangular path. $\endgroup$ – Tony Apr 17 '16 at 19:23
  • $\begingroup$ I showed you that the Hankel contour becomes a rectangular contour after a change of variable : they are the same $\endgroup$ – reuns Apr 17 '16 at 19:35
  • $\begingroup$ I understand and already expected that since the rectangular contour is a deformation of the Hankel contour which does not run into any poles. The issue I am having is that I was instructed to use the Hankel type contour because it would be easiest. $\endgroup$ – Tony Apr 18 '16 at 3:25
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If you want to use the particular keyhole contour you have in mind, you can approximate it as follows. Let $\gamma$ be a parametrization of that contour and $\phi$ be the angle the keyhole meets the lines of the contour. Then:

$$ \int_{\gamma}=\lim_{\substack{M\to\infty\\\epsilon\to 0\\\phi\to 0}}\int_{-M}^{\epsilon\cos(\phi)}+\lim_{\substack{\epsilon\to 0\\\phi\to 0}}\int_{-\pi+\phi}^{\pi-\phi}+\lim_{\substack{M\to\infty\\\epsilon\to 0\\\phi\to 0}}\int_{\epsilon\cos(\phi)}^{-M} $$

The first and third integrals cancel as usual because of the reversed direction, so you are left only with the middle term, for which you use the usual parametrization: $t=\epsilon\cdot e^{i\theta}$. Then the term becomes:

$$ \int_{-\pi+\phi}^{\pi-\phi}\frac{\theta\epsilon e^{i\theta}}{1-e^{e^{i\theta}}}d\theta $$

Now get rid of the singularity in the denominator, using the expansion:

$$1-e^t=t-\frac{1}{2}t^2+\frac{1}{3!}t^3+O(t^4)$$

Substitute $t=\epsilon e^{i\theta}$, to get:

$$\int_{-\pi+\phi}^{\pi-\phi}\frac{\theta\epsilon e^{i\theta}}{1-e^{e^{i\theta}}}d\theta=\int_{-\pi+\phi}^{\pi-\phi}\frac{\theta}{1-\frac{1}{2}\epsilon e^{i\theta}+\frac{1}{3!}(\epsilon e^{i\theta})^2\pm O(\epsilon^n)}d\theta$$

Notice that no matter the $\theta$, all exponentials in the series are bounded, so you can approximate the integral for the upper part of the circular part of the contour, as:

$$ \begin{align} E_U=\left|\int_0^{\pi-\phi}\frac{\theta\epsilon e^{i\theta}}{1-e^{e^{i\theta}}}d\theta\right|&\le \int_0^{\pi-\phi}\left|\frac{\theta}{1-O(\epsilon^n)}\right|d\theta\\ &\le\frac{1}{|1-O(\epsilon^n)|}\int_0^{\pi-\phi}|\theta| d\theta\\ &=\frac{1}{2}\cdot\frac{(\pi-\phi)^2}{|1-O(\epsilon^n)|}\\ &\to \lim_{\substack{\epsilon\to 0\\\phi\to 0}}\frac{1}{2}\cdot\frac{(\pi-\phi)^2}{|1-O(\epsilon^n)|}\\ &=\frac{\pi^2}{2} \end{align} $$

Similarly, for the lower part of the circular contour:

$$E_L=\left|\int_{-\pi+\phi}^0\frac{\theta\epsilon e^{i\theta}}{1-e^{e^{i\theta}}}d\theta\right|\le \int_{-\pi+\phi}^0\left|\frac{\theta}{1-O(\epsilon^n)}\right|d\theta\to\lim_{\substack{\epsilon\to 0\\\phi\to 0}}\frac{1}{2}\cdot\frac{(-\pi+\phi)^2}{|1-O(\epsilon^n)|}=\frac{\pi^2}{2}$$

Therefore the integral of the circular part of the contour is always bounded by the difference $E_U-E_L$, which is 0, and your result follows.

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  • $\begingroup$ Not quite what I was looking for but correct nonetheless. Thanks. $\endgroup$ – Tony Apr 20 '16 at 6:22

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