Is there a closed form for $\int_0^\infty\frac{\tanh^3(x)}{x^2}dx$?

Question

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

$$ \begin{align} I_{2,2}&= 2e_3 \\ \\ I_{4,2}&= \dfrac83e_3-4e_5 \\ \\ I_{4,4}&= -\dfrac{16}{3}e_5+20e_7 \\ \\ I_{6,2}&=\dfrac{46}{15}e_3-8e_5+6e_7 \\ \\ I_{6,4}&=-\dfrac{92}{15}e_5+40e_7-56e_9 \\ \\ I_{6,6}&=\dfrac{46}{5}e_7-112e_9+252e_{11} \\ \\ I_{3,3}&= -e_3+6e_5 \\ \\ I_{5,3}&= -e_3+10e_5-15e_7 \\ \\ I_{5,5}&= e_5-25e_7 +70e_9 \\ \\ &etc. \end{align}$$

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 used here 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.

Answer 1

Following the suggestion I made in a comment, the integral can be rewritten as the contour integral $$ I_{3,2} = \frac{1}{2\pi i} \oint \frac{\operatorname{tanh}^3 z}{z^2} \log(-z) \, dz , $$ 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.

Edit: Using $\gamma =12\,\log(A)-\log(2\pi)+\frac{6}{\pi^2}\,\zeta'(2)$, this can be simplified to $$I_{3,2}=\frac{5}{6} - \frac{4\log 2}{15} -\frac{6\zeta'(2)}{\pi^2}+ \frac{90 \zeta'(4)}{\pi^4},$$ that is $$\boxed{I_{3,2}=\frac{5}{6} - \frac{4\log 2}{15} -\frac{\zeta'(2)}{\zeta(2)}+ \frac{\zeta'(4)}{\zeta(4)}}. $$ And this form may hint to the existence of similar closed forms for other integrals $I_{n,m}$ with $n+m$ odd.

Edit: Indeed... Numerically, it looks like e.g. $$ I(5,2)=\frac1{3}\Bigl(\frac{8}{5}-\frac{44}{63}\log2-3\frac{\zeta'(2)}{\zeta(2)} +5 \frac{\zeta'(4)}{\zeta(4)}-2\frac{\zeta'(6)}{\zeta(6)} \Bigr)$$ $$ I(7,2)=\frac1{45}\Bigl(\frac{7943}{420}-\frac{428}{45}\log2- 45\frac{\zeta'(2)}{\zeta(2)}+ 98\frac{\zeta'(4)}{\zeta(4)}- 70\frac{\zeta'(6)}{\zeta(6)}+ 17\frac{\zeta'(8)}{\zeta(8)}\Bigr)$$ $$ I(9,2)=\frac1{315}\Bigl(\frac{71077}{630}-\frac{10196}{165}\log2- 315\frac{\zeta'(2)}{\zeta(2)}+ 818\frac{\zeta'(4)}{\zeta(4)}- 798\frac{\zeta'(6)}{\zeta(6)}+ 357\frac{\zeta'(8)}{\zeta(8)}- 62\frac{\zeta'(10)}{\zeta(10)}\Bigr)$$ $$I(11,2)=\frac1{14175}\Bigl(\frac{2230796}{495}-\frac{10719068}{4095}\log2- 14175\frac{\zeta'(2)}{\zeta(2)}+ 41877\frac{\zeta'(4)}{\zeta(4)}- 50270\frac{\zeta'(6)}{\zeta(6)}+ 31416\frac{\zeta'(8)}{\zeta(8)}- 10230\frac{\zeta'(10)}{\zeta(10)}+1382\frac{\zeta'(12)}{\zeta(12)}\Bigr)$$ $$I(13,2)=\frac1{467775}\Bigl(\frac{270932553}{2002}-\frac{25865068}{315}\log2- 467775\frac{\zeta'(2)}{\zeta(2)}+ 1528371\frac{\zeta'(4)}{\zeta(4)}- 2137564\frac{\zeta'(6)}{\zeta(6)}+ 1672528\frac{\zeta'(8)}{\zeta(8)}- 771342\frac{\zeta'(10)}{\zeta(10)}+197626\frac{\zeta'(12) }{\zeta(12)}-21844\frac{\zeta'(14) }{\zeta(14)}\Bigr)$$

Here the initial denominators are chosen as the least common denominators of the $\frac{\zeta'(2k)}{\zeta(2k)}$ terms, such that inside the big parentheses, we have integer coefficients here. It turns out that the sequence of those denominators, viz. $1, 3, 45, 315, 14175, 467775$, coincides up to signs with A117972, the numerators of the rational numbers $\frac{\pi^{2n}\;\zeta'(-2n)}{\zeta(2n+1)}$.

Moreover, note that the coefficients of the $\frac{\zeta'(2k)}{\zeta(2k)}$ have alternating signs and always sum to 0, meaning that the closed forms can be decomposed into terms $\frac{\zeta'(2k)}{\zeta(2k)}-\frac{\zeta'(2k-2)}{\zeta(2k-2)}$, which are thus "exponential periods".

Further, the numerators of the $\frac{\zeta'(2n)}{\zeta(2n)}$ term of $I_{2n-1,2}$, viz. $1, -2, 17, -62, 1382, -21844$, coincide with A002430, the numerators of the Taylor series for $\tanh(x)$, which are also closely related to the rational values $\zeta(1-2n)$. All this seems rather interesting.

Answer 2

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}