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For $n\geqslant m>1$, the integral $$I_{n,m}:=\int\limits_0^\infty\dfrac{\tanh^n(x)}{x^m}dx$$ converges. If $m$ and $n$ are both even or both odd, we can use the residue theorem to easily evaluate it in terms of odd zeta values, since the integrand then is a nice even function. For example, defining $e_k:=(2^k-1)\dfrac{\zeta(k) }{\pi^{k-1}}$, we have

$$ \begin{align} I_{2,2}&= 2e_3 \\ I_{4,2}&=\dfrac43(2e_3-3e_5) \\ I_{6,2}&=\dfrac2{15}(23e_3-60e_5+45e_7) \\ I_{4,4}&=\dfrac1{3}( -16e_5+60e_7) \\ I_{6,4}&=\dfrac4{15}(-23e_5+150e_7-210e_9) \\ I_{3,3}&= -e_3+6e_5 \\ I_{5,3}&= -e_3+10e_5-15e_7 \\ I_{5,5}&= e_5-25e_7 +70e_9 \\ &etc. \end{align}$$

But:

Is there a closed form for $I_{3,2}=\int\limits_0^\infty\dfrac{\tanh^3(x)}{x^2}dx$?

I am not sure at all whether nospoon's method or one of the other ad hoc approaches can be generalized to tackle this.
If the answer is positive, there might be chances that $I_{\frac32,\frac32}$ and the like also have closed forms.

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    $\begingroup$ I don't know whether to hope that we do, or that we don't, find an integration technique called "nospoon's method" in future calculus textbooks. $\endgroup$ – LSpice Jun 6 '17 at 17:04
  • $\begingroup$ A worth noticing feature of your expressions of the considered integrals in terms of the $e_{i}$ is that they're linear, so maybe using a wise mix of integration techniques and linear algebra could lead to something interesting. $\endgroup$ – Sylvain JULIEN Jun 6 '17 at 20:21
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    $\begingroup$ $I_{3,2} \approx 1.154785313323$ is not recognized by the ISC ... isc.carma.newcatle.edu.au $\endgroup$ – Gerald Edgar Jun 7 '17 at 0:38
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    $\begingroup$ @GeraldEdgar Yes I know but wow, that "really means nothing". Note that the ISC doesn't even find $1.1274284420316\approx I_{3,3}$ :-( $\endgroup$ – Wolfgang Jun 7 '17 at 6:43
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    $\begingroup$ @user75829 Interesting. Never heard of gd(x) before! But as far as I see, it is essentially just notation? $\endgroup$ – Wolfgang Jul 6 '17 at 18:23
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Following the suggestion I made in a comment, the integral can be rewritten as the contour integral $$ I_{3,2} = \frac{1}{2\pi i} \oint \frac{\operatorname{tahn}^3 z}{z^2} \log(-z) \, dz , $$ where the clockwise contour tightly encircles the positive real axis, which coincides with the branch cut of the logarithm. The reason that this integral is equivalent is because the branch jump across the real line of $\frac{1}{2\pi i} \log(-z)$ is precisely $1$.

The integrand has poles at all $z=\pm i\pi(k+\frac{1}{2})$, $k=0,1,2,\ldots$. Evaluating the residues we find \begin{align*} I_{3,2} &= \sum_{k=0}^\infty \frac{8\log\pi(k+\frac{1}{2})}{\pi^2 (2k+1)^2} - \frac{96 \log\pi(k+\frac{1}{2})-80}{\pi^4 (2k+1)^4} \\ &= \frac{5}{6} - \gamma - \frac{19 \log 2}{15} + 12 \log A - \log\pi + \frac{90 \zeta'(4)}{\pi^4} \\ &= 1.1547853133231762640590704519415261475352370924508924890\ldots \end{align*} The last two lines can be checked with Wolfram Alpha, where $\gamma$ is the Euler-Mascheroni constant, and $A$ is the Glaisher constant.

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  • $\begingroup$ Now that is cute. Thank you very much! $\endgroup$ – Wolfgang Jun 7 '17 at 20:02
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    $\begingroup$ I wonder whether this sets a record for largest number of important constants in a single formula. $\endgroup$ – Gerry Myerson Jun 7 '17 at 23:27
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Rewrite the integrand and apply Taylor expansion to $\frac1{(1+e^{-2x})^3}$ so that $$\frac{\tanh^3x}{x^2}=\sum_{j\geq0}(-1)^j\binom{j+2}2 \frac{(1-e^{-2x})^3}{x^2}\binom{j+2}2e^{-2jx}.$$ Integrate term-wise to get (after some regrouping) $$\int_0^{\infty}\frac{\tanh^3x}{x^2}\,dx=\sum_{k=2}^{\infty}(-1)^k(8k^3+4k)\log k.$$ Perhaps there is some hope in view of what I see as $$\sum_{k=2}^{\infty}(-1)^k\log k=\log\sqrt{\frac2{\pi}};$$ which is a Wallis-type formula $$\frac23\cdot\frac45\cdot\frac67\cdots\frac{2k}{2k+1}\cdots=\sqrt{\frac2{\pi}}.$$

UPDATE. Using a divergent series approach on $\sum_k(-1)^kk^c$ and Will Sawin's comment, we can complete the solution as follows. Start with $\sum_{k\geq1}(-1)^kk^c=\zeta(-c)(2^{c+1}-1)$ to get the derivate $$\sum_{k\geq2}(-1)^ck^c\log k=-\zeta'(-c)(2^{c+1}-1)+\zeta(-c)2^{c+1}\log2.$$ Now, apply the following facts: $\zeta(-1)=-\frac1{12},\, \zeta'(-1)=\frac1{12}-\log A,\, \zeta(-3)=\frac1{120}$ and $$\zeta'(-3)=\frac1{120}\log(2\pi)-\frac{11}{720}+\frac1{120}\gamma-\frac{3\zeta'(4)}{4\pi^4}.$$ Next, put all these together and simplify \begin{align}\sum_{k\geq0}(-1)^k(8k^3+4k)\log k &=[-120\zeta'(-3)+128\zeta(-3)\log 2]+[-12\zeta'(-1)+16\zeta(-1)\log 2] \\ &=\frac56-\gamma-\log\pi-\frac{19}{15}\log2+12\log A+\frac{\zeta'(4)}{\zeta(4)}. \end{align}

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  • $\begingroup$ Interesting approach with divergent series. So the first step would be to evaluate $\sum_{k=2}^{\infty}(-1)^kk\log k=\log (\dfrac{2^2}{3^3}\dfrac{4^4}{5^5}\cdots)$. But how to regularize that between the even and odd partial products...?? $\endgroup$ – Wolfgang Jun 7 '17 at 6:56
  • $\begingroup$ One may start by finding a divergent series formula for $\sum_{k\geq2}(-1)^kk^c$ for a range of values of $c$ real. Take derivative to get $\sum_k(-1)^kk^c\log k$ and then put $c=1$. $\endgroup$ – T. Amdeberhan Jun 7 '17 at 7:17
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    $\begingroup$ $\sum_k (-1)^k k^c$ is $\left( \sum_k k^c \right) (-1 + 2^{1+c} )$ and the first term is of course $\zeta(-c)$. This is surely related to the appearance of the derivative of $\zeta$ in Igor's formula. $\endgroup$ – Will Sawin Jun 7 '17 at 19:03
  • $\begingroup$ This is super, Ramanujan style! $\endgroup$ – Lewi_Sol Jun 7 '17 at 22:48
  • $\begingroup$ Indeed very cute, as well as Igor Khavkine's approach above. $\endgroup$ – Zurab Silagadze Jun 8 '17 at 5:48

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