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Evaluate

$$\int_0^\infty\left(\frac{\tanh x}{x^2}-\frac{1}{xe^{2x}}\right)dx$$

I haven't been able to find references to the indefinite integral of the $\tanh$ term except for some similar forms that had solutions. see here and here.

Edit: Following Random Variable's result we have the form $$12\log A-\frac{4}{3}\log 2$$

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  • $\begingroup$ Do you mean $\tanh(x)?$ what is $\tanh$ being applied to? $\endgroup$ Oct 10, 2017 at 23:59
  • $\begingroup$ I relied on Wolfram Alpha to evaluate a limit, but looks like it's $$8 \psi^{-2} \left(\frac{1}{2} \right) + \log(2) -2 \log(4 \pi),$$ where $\psi^{-2} \left(\frac{1}{2} \right) = \int_{0}^{1/2} \ln \Gamma(x) \, dx$, which can be expressed in terms of the Glaisher–Kinkelin constant. $\endgroup$ Oct 11, 2017 at 3:13
  • $\begingroup$ All approaches are appreciated, but it’s understandable. And why the close votes? $\endgroup$
    – tyobrien
    Oct 11, 2017 at 17:14

3 Answers 3

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Let $$I(z) =\frac{z}{a}\int_0^\infty \left(\frac{1}{2}-\frac{z}{at}+\frac{1}{e^{at/z}-1}\right) \frac{1-e^{-at}}{t^2} dt$$ Then Binet's first formula says $$I(z) = \int_0^z \left[\ln\Gamma(x) - (z-\frac{1}{2})\ln z + z - \frac{\ln(2\pi)}{2}\right] dx $$


Letting $a=2,z=1/2$ gives $$4I(\frac{1}{2}) =\int_0^\infty \left(\frac{1}{2}-\frac{1}{4t}+\frac{1}{e^{4t}-1}\right) \frac{1-e^{-2t}}{t^2} dt$$ and $a=4,z=1$ gives $$4I(1) =\int_0^\infty \left(\frac{1}{2}-\frac{1}{4t}+\frac{1}{e^{4t}-1}\right) \frac{1-e^{-4t}}{t^2} dt$$

Some algebraic manipulation yields $$4I(1)-4I(\frac{1}{2}) = \underbrace{\int_0^\infty \left[\frac{1}{2t^2}-\frac{e^{-2t}}{2t} - (\frac{1}{2}-\frac{1}{4t})\left(\frac{e^{-2t}-e^{-4t}}{t^2}\right)\right] dt}_{J} - \frac{I}{2}$$

With $I$ your desired integral. Surprisingly, $J$ has elementary primitive (even it were not, we still have systematic way to crash it), with value $5/8$.


Hence it remains to evalute $$\int_0^1 \ln \Gamma(x) dx \quad \quad \int_0^{1/2} \ln \Gamma(x) dx$$ The former is just $\ln(2\pi)/2$, for the latter, we can use the integral representation of Barnes G function: $$\int_0^z \ln \Gamma(x) dx = \frac{z(1-z)}{2}+\frac{z}{2}\ln(2\pi) + z \ln\Gamma(z) - \log G(1+z)$$

and the special value $$\ln G(\frac{3}{2}) = -\frac{3}{2}\ln A + \frac{\ln \pi}{4}+\frac{1}{8}+\frac{\ln 2}{24}$$

with $A$ being the Glaisher-Kinkelin constant.

Alternatively, use Fourier expansion of $\ln \Gamma(x)$, integrate termwise, and remember the relation between $A$ and $\zeta'(2)$ also gives the value of the integral $$\int_0^{1/2} \ln\Gamma(x)dx = \frac{3}{2}\ln A + \frac{5}{24}\ln 2 + \frac{\ln \pi}{4}$$

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Let $$I(a,b) = \int_{0}^{\infty} \left(\frac{\tanh (x)}{x} -e^{-bx} \right)\frac{e^{-ax}}{x} \, dx, $$ where $a, b >0$.

I will show that $$I(a,b) = 8 \psi^{(-2)} \left(\frac{a+2}{4} \right)- 8 \psi^{(-2)}\left(\frac{a}{4} \right)-a \log(a)+a + 2 a \log (2)+ \log(a+b)- 2 \log(4 \pi),$$ where $\psi^{(-2)}(x)$ is the polygamma function of order $-2$ defined by the integral $$ \psi^{(-2)}(x) = \int_{0}^{x} \log \Gamma (t) \, dt.$$

If we let $a \to 0^{+}$, we get $$\int_{0}^{\infty} \left(\frac{\tanh (x)}{x} -e^{-bx} \right)\frac{dx}{x} = 8 \psi^{(-2)} \left(\frac{1}{2} \right)+ \log(b) - 2 \log(4\pi). $$

As explained in pisco's answer, $\psi^{(-2)} \left(\frac{1}{2} \right)$ can be expressed in terms of the Glaisher-Kinkelin constant.


Differentiating $I(a,b)$ under the integral sign with respect to $a$ (which is permissible for $a \ge c$, where $c$ is some positive value), we get $$ \frac{\partial}{\partial a}I(a,b) = -\int_{0}^{\infty}\left(\frac{\tanh (x)}{x}-e^{-bx} \right) e^{-ax} \, dx = -\int_{0}^{\infty} \frac{\tanh (x)}{x} \, e^{-ax} \, dx + \frac{1}{a+b}.$$

And using a property of the Laplace transform, we get

$$\begin{align}\int_{0}^{\infty} \frac{\tanh (x)}{x} \, e^{-ax} \, dx &= \int_{a}^{\infty} \int_{0}^{\infty} \tanh (x) e^{-px} \, dx \, dp \\ &=\int_{a}^{\infty} \int_{0}^{\infty} \frac{1-e^{-2x}}{1+e^{-2x}} \, e^{-px} \, dx \, dp \\ &= \int_{a}^{\infty} \int_{0}^{\infty} \left(2\sum_{n=0}^{\infty} (-1)^{n} e^{-2nx} -1\right) e^{-px} \, dx \, dp \\ &= \int_{a}^{\infty} \left( 2 \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+p}- \frac{1}{p}\right) \, dp \\ &= \int_{a}^{\infty} \left(\frac{2}{p} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{\frac{2}{p}n +1} - \frac{1}{p}\right) \, dp \\ &= \int_{a}^{\infty} \left(\frac{1}{2} \left(\psi \left(\frac{p+2}{4} \right)- \psi\left(\frac{p}{4} \right)\right)- \frac{1}{p} \right) \, dp \tag{1} \\ &= -2 \log (2) -2\log \Gamma \left(\frac{a+2}{4} \right) +2 \log \Gamma \left(\frac{a}{4} \right) + \log (a). \tag{2} \end{align}$$

Therefore, $$I(a,b) =2a \log(2) + 8 \psi^{(-2)} \left(\frac{a+2}{4} \right)- 8 \psi^{(-2)} \left(\frac{a}{4} \right)- a \log(a) +a + \log(a+b) +C. $$

To determine the integration constant $C$, we can take the limit on both sides of the above equation as $a \to +\infty$.

Since $\psi^{(-2)}(x) $ can be expressed in terms of the Barnes G-function, we can use the asymptotic expansion of the Barnes G-function.

(Term-by-term integration of Stirling's formula for the log gamma function does lead to the same expansion but with an unknown constant.)

The integration constant $C$ turns out to be $-2 \log(4 \pi)$.


$(1)$ https://mathworld.wolfram.com/DigammaFunction.html (6)

$(2)$ Using Stirling's formula, we have

$$\log \Gamma\left(\frac{a+2}{4} \right) \sim \left(\frac{a+2}{4} - \frac{1}{2} \right) \log \left(\frac{a+2}{4} \right) - \frac{a+2}{4} + \frac{1}{2} \log(2 \pi) + \mathcal{O}\left(\frac{1}{a} \right), $$ where

$$\log \left(\frac{a+2}{4} \right)= \log \left(\frac{a}{4} \right) + \log \left(1+ \frac{2}{a} \right) \sim \log \left(\frac{a}{4} \right) + \frac{2}{a} + \mathcal{O}\left(\frac{1}{a^{2}} \right).$$

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$(A)\enspace$ Dissolving the integral

With the split

$\displaystyle \frac{\tanh x}{x^2}-\frac{1}{x e^{2x}} =$

$\displaystyle =\left( \frac{2}{e^{2x}} -\frac{1}{e^{4x}} -\frac{1}{e^x} \right) \frac{1}{x} - \frac{e^{2x}-1}{e^{4x} (e^{2x}+1)} \frac{1}{x} + \frac{1}{e^{2x}+1} \frac{e^{2x}-1-2x}{x^2} + \frac{1}{x e^x} - \frac{1}{x e^{2x}} $

follows:

$\displaystyle \int\limits_0^\infty \left(\frac{\tanh x}{x^2}-\frac{1}{x e^{2x}}\right) dx = \int\limits_0^\infty \left( \frac{2}{e^{2x}} -\frac{1}{e^{4x}} -\frac{1}{e^x} \right) \frac{dx}{x} - \int\limits_0^\infty \frac{e^{2x}-1}{e^{4x} (e^{2x}+1)} \frac{dx}{x} + $

$\hspace{5cm}\displaystyle +\int\limits_0^\infty \frac{1}{e^{2x}+1} \sum\limits_{n=2}^\infty \frac{2^n x^{n-2}}{n!} dx + \int\limits_0^\infty \left(\frac{1}{x e^x} - \frac{1}{x e^{2x}}\right) dx =$

$\displaystyle = \text{Ei}(0)\cdot (2-1-1) - \int\limits_0^\infty \frac{(1-e^{-ax})(1-e^{-bx})}{x(e^x-1)}dx |_{a=b=\frac{1}{2}} + \sum\limits_{n=2}^\infty \frac{2^n}{n!} \int\limits_0^\infty \frac{x^{n-2}}{e^{2x}+1} dx + \ln 2 $

$\displaystyle = 0 + \ln\frac{\Gamma(1+a)\Gamma(1+b)}{\Gamma(1+a+b)} |_{a=b=\frac{1}{2}} + \sum\limits_{k=2}^\infty \frac{2^k}{k!}\eta(k-1)\Gamma(k-1)2^{1-k} + \ln 2 $

$\displaystyle = 2\ln2 + \ln\frac{\pi}{4} + 2 \sum\limits_{n=2}^\infty \frac{\eta(n)}{n(n+1)} $


$(B)\enspace$ Basics

The following calculation of $\enspace\displaystyle \sum\limits_{n=2}^\infty \frac{\eta(n)}{n(n+1)}\enspace $ is based on a generalization of the Gamma function,

called $Q_m(x)$ in https://www.fernuni-hagen.de/analysis/download/bachelorarbeit_aschauer.pdf ,

with $\enspace\displaystyle \ln Q_m(x)=\frac{(-x)^{m+1}}{m+1}\gamma +\sum\limits_{n=2}^\infty\frac{(-x)^{m+n}}{m+n}\zeta(n)\enspace$ on page $13$, $(4.2)\,$ .

A modification is $\enspace\displaystyle Q_m^*(x):=(1+x)Q_m(x)\enspace$ on page $24$, $(4.8)$ and $(4.9)$ .

Special values which are used here are $\,Q_0^*(-1)=1\,$, $\,\displaystyle Q_1^*(-1)=\frac{1}{\sqrt{2\pi}}\,$ ,

$\displaystyle Q_0(-\frac{1}{2})=\Gamma(\frac{1}{2})=\sqrt{\pi}\,$ and $\,\displaystyle Q_1(-\frac{1}{2})=A^{\frac{3}{2}}2^{-\frac{7}{24}}\,$ (page $38$) .

With the generalized Glaisher-Kinkelin constant $\,A_n\,$ it’s $\,A:=A_1\,$ .

Note: $\enspace\displaystyle Q_0(x)=\Gamma(1+x)\,$ , $\enspace\displaystyle Q_1(x)=\frac{\sqrt{2\pi}^x}{e^{\frac{1}{2}x(x+1)}G(1+x)}\,$ ($G$ is the Barnes G function)

$\hspace{1.4cm}$Special value: $\enspace\displaystyle G(\frac{1}{2})=2^\frac{1}{24}e^\frac{1}{8}\pi^{-\frac{1}{4}}A^{-\frac{3}{2}}$


$(C)\enspace$ Calculating the series

$\displaystyle \sum\limits_{n=2}^\infty \frac{x^n}{n(n+1)} \eta(n) =$

$\displaystyle =\left(\sum\limits_{n=2}^\infty \frac{x^n}{n} \zeta(n) - \sum\limits_{n=2}^\infty \frac{x^n}{n+1} \zeta(n)\right) - 2 \left(\sum\limits_{n=2}^\infty \frac{x^n}{2^n n} \zeta(n) - \sum\limits_{n=2}^\infty \frac{x^n}{2^n (n+1)} \zeta(n)\right) $

$\displaystyle = \left(\ln Q_0(-x) - x\gamma - \frac{1}{x}(\ln Q_1(-x)-\frac{x^2}{2}\gamma) \right) $

$\hspace{0.5cm}\displaystyle - 2 \left(\ln Q_0(-\frac{x}{2} ) - \frac{x}{2}\gamma - \frac{2}{x}(\ln Q_1(-\frac{x}{2})-\frac{x^2}{8}\gamma) \right) $

$\displaystyle = \ln Q_0^*(-x) - \frac{1}{x}\ln Q_1^*(-x) + \frac{1-x}{x}\ln(1-x) - 2 \ln Q_0(-\frac{x}{2}) + \frac{4}{x}\ln Q_1(-\frac{x}{2}) $

With $\,x\uparrow 1\,$ follows

$ \displaystyle \sum\limits_{n=2}^\infty \frac{\eta(n)}{n(n+1)} = \ln Q_0^*(-1) - \ln Q_1^*(-1) - 2 \ln Q_0(-\frac{1}{2}) + 4 Q_1(-\frac{1}{2}) $

$\hspace{2.5cm}\displaystyle = -\frac{2}{3}\ln 2 - \frac{1}{2}\ln \pi + 6 \ln A $

and therefore

$\displaystyle \int\limits_0^\infty \left(\frac{\tanh x}{x^2}-\frac{1}{x e^{2x}}\right) dx = 2\ln2 + \ln\frac{\pi}{4} + 2 \left(-\frac{2}{3}\ln 2 - \frac{1}{2}\ln \pi + 6 \ln A \right) $

$\hspace{4.5cm}\displaystyle = 12 \ln A - \frac{4}{3}\ln 2 \enspace$ .

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