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Prove that $$ \int_{-\infty}^\infty \frac{1}{(\coth(x)+x)^2+\frac{\pi^2}{4}} \mathrm{d}x =1 $$

The hyperbolic function $\coth x$ is defined by: $$ {\displaystyle \coth x={\frac {\cosh x}{\sinh x}}={\frac {e^{x}+e^{-x}}{e^{x}-e^{-x}}}={\frac {e^{2x}+1}{e^{2x}-1}}}. $$

I got this problem from my friend and I tried hard to solve it but I don't know how to begin with it: all I can observe is that the denominator can never be zero. A similar problem I found in this site is this:

Prove this $ \int_0^\infty\frac{\coth^2x-1}{(\coth x-x)^2+\frac{\pi^2}{4}}dx=\frac45 $

Could anyone please help?

Ahmad Bow
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    As this is the third question in a row posted by you, just stating a problem and merely showing what you've tried, I'll point out this issue here: MSE encourages to not just state questions but also to include how you approached to them and what you really tried. Simply saying "I tried hard to solve" isn't enough; what exactly did you tried? You were using the tag complex-analysis for every of the aforementioned questions too, but there was no sign of complex analysis used... Show, don't tell. Include where you started, which road you took and where precisely you got stuck. – mrtaurho Oct 27 '19 at 18:22
  • The only look at the question gives me a feeling that somehow Glasser's Master theorem can be applied here.... – Rohan Shinde Oct 27 '19 at 18:33
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    You might want to take a look here:https://math.stackexchange.com/questions/1015462/a-strange-integral-int-infty-infty-dx-over-1-leftx-tan-x-rig – Rohan Shinde Oct 27 '19 at 18:39
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    @mrtaurho why dont we focus on how beautiful the question is and how can we approach it instead of focusing on what he did or did not try and away from the policy of this site? For me, I dont care if the OP tried it or not as long as the problem is nice and if I dont like the problem or too hard for me to solve I just simply ignore because it could be a nice problem for others. – Ali Shadhar Oct 27 '19 at 18:49
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    @AliShather I'm only trying to explain our MSE policy to a user consequently ignoring it. Don't get me wrong, I'm not saying I completely agree with this philosophy of responding to newly posted question, but we've rules for a reason. In general a question without any attempt shown could be a homework assignment and answering such a question doesn't seem reasonable for me, does it to you? To me it looks like "Here, I've got a task, I don't can (or want) to solve. Do it for me cause I've to hand it in tomorrow!" and I dislike such posts. – mrtaurho Oct 27 '19 at 18:57
  • Totally makes sense and you provided a nice example.. but for me I would never down vote or close vote a problem I dont like whatever the reason is. – Ali Shadhar Oct 27 '19 at 19:01
  • @Ahmad Bow just post any question you like and do not care about the rules of this site or the close votes. real mathematicians would appreciate your problems/ solutions. mrtaurho gave a good example why you should show some work but you do not have to. I saw tons of questions posted without showing any work and they still there open. I am a big fan of your problems .. nice and unique. keep it up.. – Ali Shadhar Oct 27 '19 at 19:16

1 Answers1

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We will adopt the standard technique (see this.2.5 or this). Let $a > 0$ and consider the function

$$ \newcommand{\Log}{\operatorname{Log}} f(z) = \frac{1}{\frac{1}{2}\Log(-z) + a \frac{z+1}{z-1}} \cdot \frac{1}{z}, $$

where $\Log$ is the principal logarithm. Then using the substitution $x=\log\sqrt{y}$ and the keyhole contour,

\begin{align*} \int_{-\infty}^{\infty} \frac{1}{(x + a \coth(x))^2 + \frac{\pi^2}{4}} \cdot \mathrm{d}x &= \int_{0}^{\infty} \frac{1}{\bigl(\frac{1}{2}\log y + a \frac{y+1}{y-1} \bigr)^2 + \frac{\pi^2}{4}} \cdot \frac{\mathrm{d}y}{2y}\\ &= \frac{1}{2\pi i} \biggl( \int_{i0^{+}}^{+\infty+i0^{+}} f(z) \, \mathrm{d}z - \int_{-i0^{+}}^{+\infty-i0^{+}} f(z) \, \mathrm{d}z \biggr) \\ &= \sum_{z \in \mathbb{C}\setminus[0,\infty)} \operatorname{Res}(f, z). \end{align*}

But it turns out that $f$ has a unique simple pole at $z = -1$ on $\mathbb{C}\setminus[0,\infty)$, and so, the above residue is easily computed as

\begin{align*} \int_{-\infty}^{\infty} \frac{1}{(x + a \coth(x))^2 + \frac{\pi^2}{4}} \cdot \mathrm{d}x = \frac{2}{a+1}. \end{align*}

OP's question corresponds to the special case $a = 1$, proving that the answer is indeed $1$.

Sangchul Lee
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