Conjectured value of $\int_{0}^{\infty}\left(\frac{x-1}{\ln^2 x}-\frac{1}{\ln x}\right)\frac{\mathrm{d}x}{x^2+1}$

Question

I was curious whether this integral has a closed form expression :

$$\int_{0}^{\infty}\left(\frac{x-1}{\ln^2 x}-\frac{1}{\ln x}\right)\frac{\mathrm{d}x}{x^2+1}$$

The integrand has a singularity at $x=1$, but it's removable. And as $x \to \infty$, the integrand behaves like $\frac{1}{x \ln^{2}x}$. So the integral clearly converges.

Although I have not been able to derive its closed form, I think, by reverse symbolic calculators, up to 20 digits it could be

$$I=\frac{4G}{\pi}$$

where $G$ is Catalan's constant. Is it true or is it completely fabulous?

EDIT. NOTE :

For better search to this integral I have renamed the title from Conjectured value of logarithmic definite integral, which is ambiguous and did not say anything, to the current one with integral explicitly written.

Answer 1

It is not necessary to exploit any symmetries of the integrand. Setting $x=e^y$

$$ I=\int_{-\infty}^{\infty}\underbrace{e^y\left(\frac{e^y-1}{y^2}-\frac{1}{y}\right)\frac{1}{e^{2y}+1}}_{f(y)}\,dy $$

Integrating around a big semicircle in the UHP (exercise: show convergence in this domain of the complex plane) we obtain

$$ I=2 \pi i \sum_{n=0}^{\infty}\text{Res}(f(z),z=z_n) $$ here $z_n=\frac{i\pi}2(2n+1)$. This is easily rewritten as

$$ I=2 \pi i\left(\left(\frac{1}{\pi}\color{blue}{\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}}-\frac{2}{\pi^2}\color{red}{\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}}\right) -\frac{2i}{\pi^2}\color{green}{\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}}\right) $$

since $\color{blue}{\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}=\frac{\pi}{4}}$ and $\color{red}{\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}=\frac{\pi^2}{8}}$ the imaginary parts cancel and we are left with

$$ I= \frac{4}{\pi}\color{green}{\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}}=\frac{4\color{green}{K}}{\pi} $$

Answer 2

Our integral equals

$$ I=\int_{-\infty}^{+\infty}\left(\frac{e^t-1-t}{t^2}\right)\frac{e^t}{e^{2t}+1}\,dt $$ that by exploiting symmetry becomes $$ I = \int_{0}^{+\infty}\frac{e^{t}+e^{-t}-2}{t^2(e^{t}+e^{-t})}\,dt =\int_{0}^{+\infty}\frac{\cosh(t)-1}{t^2\cosh(t)}\,dt$$ The last integral is straightforward to compute trough the residue theorem. Since $$ \text{Res}\left(\frac{\cosh(t)-1}{t^2\cosh(t)},t=\frac{\pi(2k+1)}{2}i\right)= (-1)^{k+1}\frac{4i}{\pi^2(2k+1)^2}$$ we have: $$\boxed{ I = \frac{4}{\pi}\sum_{k\geq 0}\frac{(-1)^k}{(2k+1)^2}=\color{red}{\frac{4G}{\pi}}}$$

as conjectured.

Answer 3

Here is yet another approach. We first note that we can write $\frac{x-1}{\log(x)}$ as

$$\frac{x-1}{\log(x)}=\int_0^1 x^t\,dt$$

Therefore, we can write

$$\begin{align} \int_0^\infty \left(\frac{x-1}{\log^2(x)}-\frac{1}{\log(x)}\right)\frac{1}{1+x^2}\,dx&=\int_0^\infty \int_0^1 \frac{x^t-1}{\log(x)}\,\frac{1}{1+x^2}\,dt\,dx\\\\ &=\int_0^1 \int_0^\infty \frac{x^t-1}{\log(x)}\,\frac{1}{1+x^2}\,dx\,dt\tag1 \end{align}$$

Let $I(t)$ represent the inner integral of the right-hand side of $(1)$. Then, differentiating, we find that

$$\begin{align} I'(t)&=\int_0^\infty \frac{x^t}{1+x^2}\,dx\\\\ &=\frac{\pi}{2\cos(\pi t/2)}\tag 2 \end{align}$$

where I derived the right-hand side of $(2)$ in THIS ANSWER. Alternatively, using real analysis only, we have

$$\begin{align} \int_0^\infty \frac{x^t}{1+x^2}\,dx&=\frac12 B\left(\frac{1+t}{2},\frac{1-t}{2}\right)\\\\ &=\frac12 \Gamma\left(\frac{1+t}{2}\right)\Gamma\left(\frac{1-t}{2}\right)\\\\ &=\frac12\frac{\pi}{\sin\left(\pi\frac{1+t}{2}\right)}\\\\ &=\frac{\pi}{2\cos(\pi t/2)} \end{align}$$

Integrating $(2)$ and using $I(0)=0$ reveals

$$I(t)=\int_0^t \frac{\pi}{2\cos(\pi t'/2)}\,dt' \tag 3$$

Substituting $(3)$ into $(1)$ yields

$$\begin{align} \int_0^\infty \left(\frac{x-1}{\log^2(x)}-\frac{1}{\log(x)}\right)\frac{1}{1+x^2}\,dx&=\frac{\pi}{2}\int_0^1 \int_0^t \sec(\pi t'/2)\,dt'\,dt \tag 4\\\\ &=\frac{\pi}{2}\int_0^1 (1-t)\sec(\pi t/2)\,dt \tag5\\\\ &=\frac{\pi}{2}\int_0^1 t\csc(\pi t/2)\,dt \tag 6\\\\ &=\frac{1}{\pi}\int_{-\pi/2}^{\pi/2}\frac{t}{\sin(t)}\,dt \tag 7\\\\ &=\frac{4G}{\pi} \tag 8 \end{align}$$

as was to be shown!

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NOTES:

In going from $(4)$ to $(5)$, we changed the order of integration and carried out the inner integral.

In going from $(5)$ to $(6)$, we enforced the substitution $t \to 1-t$.

In going from $(6)$ to $(7)$, we enforced the substitution $t \to 2t/\pi$ and exploited the evenness of the integrand.

In going from $(7)$ to $(8)$, we made use of one of the integral identities for Catalan's Constant as found HERE.

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ALTERNATIVE DEVELOPMENT

Note that we can write $(3)$ as

$$I(t)=\log\left(\cot\left(\frac{\pi}{4}(1-t)\right)\right) \tag 9$$

Then, substituting $(9)$ into $(1)$ yields

$$\begin{align} \int_0^\infty \left(\frac{x-1}{\log^2(x)}-\frac{1}{\log(x)}\right)\frac{1}{1+x^2}\,dx&=\int_0^1 \log\left(\cot\left(\frac{\pi}{4}(1-t)\right)\right)\,dt \\\\ &=\frac{4}{\pi}\int_0^{\pi/4} \log(\cot(t))\,dt \tag 9\\\\ &=\frac{4G}{\pi} \end{align}$$

which uses another well-known integral identity for $G$ as found HERE.

Note that if we enforce the substitution $t\to \text{arccot}(t)$ in $(9)$, we find the result in terms of the series representation of $G$ as

$$\begin{align} \int_0^\infty \left(\frac{x-1}{\log^2(x)}-\frac{1}{\log(x)}\right)\frac{1}{1+x^2}\,dx&=\frac{4}{\pi}\int_0^{\pi/4} \log(\cot(t))\,dt \\\\ &=\frac{4}{\pi}\int_1^{\infty}\frac{\log(t)}{1+t^2}\,dt\\\\ &=-\frac{4}{\pi}\int_0^1 \frac{\log(t)}{1+t^2}\,dt\\\\ &=-\frac{4}{\pi}\sum_{n=0}^\infty(-1)^n \int_0^1 t^{2n}\log(t)\,dt\\\\ &=\frac{4}{\pi}\sum_{n=0}^\infty(-1)^n \int_0^1 \frac{t^{2n}}{2n+1}\,dt\\\\ &=\frac{4}{\pi}\sum_{n=0}^\infty(-1)^n \frac{1}{(2n+1)^2}\\\\ &=\frac{4G}{\pi} \end{align}$$

as expected once again!

Answer 4

Though using the residue method is somewhat straightforward, but not everyone can understand it. So, here is a residue-free method:

Split the integral into two terms where each term is in the interval $0<x<1$ and $1<x<\infty$, then use the substitution $x\mapsto\frac{1}{x}$ to the second term. We will get $$ \left[\int_{0}^{1}+\int_{1}^{\infty}\right]\left(\frac{x-1}{\ln^2 x}-\frac{1}{\ln x}\right)\frac{\mathrm{d}x}{x^2+1}=\int_{0}^{1}\frac{(x-1)^2}{x\ln^2 x}\cdot\frac{\mathrm{d}x}{x^2+1}\tag1 $$ Now, for $a\ge-1$ , one may consider the following integral $$ I(a)=\int_{0}^{1}x^a\cdot\frac{(x-1)^2}{\ln^2 x}\cdot\frac{\mathrm{d}x}{1+x^2}\tag2 $$ and the desired integral is $I(-1)$. Since $0<x<1$, one may observe that $I(\infty)\to0$ as $a\to\infty$. \begin{align} I''(a)&=\int_{0}^{1}\frac{x^a(x-1)^2}{1+x^2}\ \mathrm{d}x\\[10pt] &=\int_{0}^{1}\sum_{k=0}^\infty(-1)^k\ x^{2k+a}\ (x^2-2x+1)\ \mathrm{d}x\\[10pt] &=\sum_{k=0}^\infty(-1)^k\left(\frac{1}{2k+a+3}-\frac{2}{2k+a+2}+\frac{1}{2k+a+1}\right)\\[10pt] &=\frac{1}{4}\left[\psi\left(\frac{a+5}{4}\right)-2\psi\left(\frac{a+4}{4}\right)+2\psi\left(\frac{a+2}{4}\right)-\psi\left(\frac{a+1}{4}\right)\right]\\[10pt] I'(a)&=\ln\Gamma\left(\frac{a+5}{4}\right)-\ln\Gamma\left(\frac{a+1}{4}\right)+2\ln\Gamma\left(\frac{a+2}{4}\right)-2\ln\Gamma\left(\frac{a+4}{4}\right)\\[10pt] I(a)&=4\left[\psi\left(-2,\frac{a+5}{4}\right)-\psi\left(-2,\frac{a+1}{4}\right)+2\psi\left(-2,\frac{a+2}{4}\right)-2\psi\left(-2,\frac{a+4}{4}\right)\right]\tag3\\[10pt] \end{align} Hence $$ I(-1)=4\left[\psi\left(-2,1\right)-\psi\left(-2,0\right)+2\psi\left(-2,\frac{1}{4}\right)-2\psi\left(-2,\frac{3}{4}\right)\right]=\frac{4G}{\pi} $$ Wolfram Alpha confirms it. One may also use the special values of generalized polygamma function and its related relation with derivative of Hurwitz Zeta Function: $$\psi(-2,x)=\zeta'(-1,x)-\frac{x^2}{2}+\frac{x}{2}-\frac{1}{12}$$

Answer 5

Jack D'Aurizio showed that $$\int_{0}^{\infty}\left(\frac{x-1}{\ln^2 x}-\frac{1}{\ln x}\right)\frac{dx}{x^2+1} = \int_{0}^{\infty} \left( 1-\frac{1}{\cosh x} \right) \frac{dx}{x^{2}} .$$

The following is an alternative evaluation of the integral on the right.

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An integral representation of the Dirichlet beta function is $$\beta(s) = \frac{1}{ 2 \, \Gamma(s)} \int_{0}^{\infty} \frac{x^{s-1}}{\cosh(x)} \, dx \, , \quad \text{Re}(s) >0\tag{1}. $$

And the Laplace transform of $x^{s-1}$ is $$\int_{0}^{\infty} x^{s-1} e^{-ax} \, dx = \frac{\Gamma(s)}{a^{s}} \, , \quad (\text{Re}(s) >0, \ \text{Re}(a)>0) \tag{2}.$$

Subtracting $(1)$ from $(2)$, we get $$\int_{0}^{\infty}\left(e^{-ax} - \frac{1}{\cosh (x)} \right) x^{s-1} \, dx = \Gamma(s) \left( a^{-s} - 2 \beta(s) \right) , \tag{3}$$ which holds for $ \text{Re}(s) > -1$ and $\text{Re}(a) > 0$.

If we restrict $s$ to that strip $-1 < \text{Re}(s) <0$, then $(3)$ also holds for $a = 0$.

From the functional equation of the Dirichlet beta function, we see that the Dirichlet beta function has a zero at $s=-1$.

So letting $s$ tend to $-1$, we get

$$ \begin{align} \int_{0}^{\infty} \left(1- \frac{1}{\cosh x} \right) \frac{dx}{x^{2}} &= \lim_{s \downarrow -1} \Gamma(s) \left((0 - 2 \beta(s)\right) \\ &= - 2 \lim_{s \downarrow -1} \Gamma(s) \beta(s) \\ &=-2 \lim_{s \downarrow -1} \left(-\frac{1}{s+1} + \mathcal{O}(1) \right) \beta(s) \\ &= 2 \beta'(-1). \end{align}$$

To show that $ \displaystyle \beta'(-1) = \frac{2G}{\pi} $, differentiate both sides of the functional equation, and then let $s=2$.

Answer 6

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Note that $\ds{\quad{x - 1 \over \ln\pars{x}} = \int_{0}^{1}x^{t}\,\dd t\,,\quad x \in \pars{0,1}}$.

\begin{align} &\color{#f00}{\int_{0}^{\infty}\bracks{% {x - 1 \over \ln^{2}\pars{x}} - {1 \over \ln\pars{x}}} \,{\dd x \over x^{2} + 1}} \\[5mm] = &\ \int_{0}^{1}\bracks{% {x - 1 \over \ln^{2}\pars{x}} - {1 \over \ln\pars{x}}} \,{\dd x \over x^{2} + 1} + \int_{1}^{0}\bracks{% {1/x - 1 \over \ln^{2}\pars{1/x}} - {1 \over \ln\pars{1/x}}} \,{-\,\dd x/x^{2} \over 1/x^{2} + 1} \\[5mm] = &\ \int_{0}^{1}{\pars{x - 1}^{2} \over x\ln^{2}\pars{x}}\,{\dd x \over x^{2} + 1} = \int_{0}^{1}{1 \over x\pars{x^{2} + 1}}\int_{0}^{1}x^{y}\,\dd y\int_{0}^{1}x^{z}\,\dd z\,\dd x \\[5mm] = &\ \int_{0}^{1}\int_{0}^{1}\int_{0}^{1}{x^{y + z - 1} \over x^{2} + 1} \,\dd x\,\dd y\,\dd z = \int_{0}^{1}\int_{0}^{1}\int_{0}^{1} {x^{y + z - 1} - x^{y + z + 1}\over 1 - x^{4}}\,\dd x\,\dd y\,\dd z \\[5mm] \stackrel{x^{4}\ \mapsto\ x}{=}\,\,\, &\ {1 \over 4}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1} {x^{y/4 + z/4 - 1}\,\,\, -\,\,\, x^{y/4 + z/4 - 1/2}\over 1 - x} \,\dd x\,\dd y\,\dd z \\[5mm] = &\ {1 \over 4}\int_{0}^{1}\int_{0}^{1}\bracks{% \Psi\pars{{y + z \over 4} + \half} - \Psi\pars{{y + z \over 4}}}\,\dd y\,\dd z \\[5mm] = &\ 4\int_{0}^{1/4}\int_{0}^{1/4}\bracks{% \Psi\pars{y + z + \half} - \Psi\pars{y + z}}\,\dd y\,\dd z\tag{1} \end{align}

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$\ds{\Psi}$ is the Digamma Function and we used its well known integral representation $\ds{\pars{~\gamma\ \mbox{is the}\ Euler\mbox{-}Mascheroni\ Constant~}}$ $$ \Psi\pars{z} = -\gamma + \int_{0}^{1}{1 - t^{z - 1} \over 1 - t}\,\dd t\,, \qquad\Re\pars{z} > 0 $$

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Since $\ds{\Psi\pars{z}\ \stackrel{\mbox{def.}}{=}\ \totald{\ln\pars{\Gamma\pars{z}}}{z}}$ $\ds{\pars{~\Gamma\ \mbox{is the}\ Gamma\ Function~}}$, $\ds{\pars{1}}$ is reduced to: \begin{align} &\color{#f00}{\int_{0}^{\infty}\bracks{% {x - 1 \over \ln^{2}\pars{x}} - {1 \over \ln\pars{x}}} \,{\dd x \over x^{2} + 1}} \\[5mm] = &\ 4\int_{0}^{1/4}\bracks{\ln\pars{\Gamma\pars{z + {3 \over 4}}} - \ln\pars{\Gamma\pars{z + {1 \over 4}}} - \ln\pars{\Gamma\pars{z + \half}} + \ln\pars{\Gamma\pars{z}}}\,\dd z \\[5mm] = &\ 4\int_{0}^{1}\ln\pars{\Gamma\pars{z}}\,\dd z + 8\int_{0}^{1/4}\ln\pars{\Gamma\pars{z}}\,\dd z - 8\int_{0}^{3/4}\ln\pars{\Gamma\pars{z}}\,\dd z\tag{2} \end{align}

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The $\ds{\ln\Gamma}$-integrals are evaluated $\ds{\pars{~\mbox{the first one is rather trivial and it's equal to}\ \half\,\ln\pars{2\pi}~}}$ with the identity ( $\ds{\,\mathrm{G}}$ is the Barnes-G Function ) $$ \int_{0}^{z}\ln\pars{\Gamma\pars{z}}\,\dd z = \half\,z\pars{1 - z} + \half\,\ln\pars{2\pi}z + z\ln\pars{\Gamma\pars{z}} - \ln\pars{\,\mathrm{G}\pars{1 + z}} $$ Namely, \begin{equation} \left\lbrace\begin{array}{\rcl} \ds{\int_{0}^{1}\ln\pars{\Gamma\pars{z}}} & \ds{=} & \ds{\half\,\ln\pars{2\pi}\ \mbox{because}\ \Gamma\pars{1} = \,\mathrm{G}\pars{2} = 1.} \\[3mm] \ds{\int_{0}^{1/4}\ln\pars{\Gamma\pars{z}}} & \ds{=} & \ds{{3 \over 32} + {1 \over 8}\,\ln\pars{2\pi} + {1 \over 4}\,\ln\pars{\Gamma\pars{1 \over 4}} - \ln\pars{\,\mathrm{G}\pars{5 \over 4}}} \\[1mm] & \ds{=} & \ds{{3 \over 32} + {1 \over 8}\,\ln\pars{2\pi} - {3 \over 4}\,\ln\pars{\Gamma\pars{1 \over 4}} - \ln\pars{\,\mathrm{G}\pars{1 \over 4}}} \\[3mm] \ds{\int_{0}^{3/4}\ln\pars{\Gamma\pars{z}}} & \ds{=} & \ds{{3 \over 32} + {3 \over 8}\,\ln\pars{2\pi} + {3 \over 4}\,\ln\pars{\Gamma\pars{3 \over 4}} - \ln\pars{\,\mathrm{G}\pars{7 \over 4}}} \\[1mm] & \ds{=} & \ds{{3 \over 32} + {3 \over 8}\,\ln\pars{2\pi} - {1 \over 4}\,\ln\pars{\Gamma\pars{3 \over 4}} - \ln\pars{\,\mathrm{G}\pars{3 \over 4}}} \end{array}\right.\tag{3} \end{equation} In these expressions we used $\ds{\,\mathrm{G}\pars{1 + z} = \,\mathrm{G}\pars{z}\Gamma\pars{z}}$. Fortunately, values of $\ds{\,\mathrm{G}\pars{z}}$ at $\ds{z = {1 \over 4}, {3 \over 4}}$ are known: \begin{align} \,\mathrm{G}\pars{1 \over 4} & = A^{-9/8}\,\,\Gamma^{\, -3/4}\pars{1 \over 4} \exp\pars{{3 \over 32} - {K \over 4\pi}}\tag{4} \\[5mm] \,\mathrm{G}\pars{3 \over 4} & = A^{-9/8}\,\,\Gamma^{\, -1/4}\pars{3 \over 4} \exp\pars{{3 \over 32} + {K \over 4\pi}}\tag{5} \end{align} $\ds{A}$ and $\ds{K}$ are the Glaisher-Kinkelin and the Catalan Constants, respectively. With $\ds{\pars{4}\ \mbox{and}\ \pars{5}}$, $\ds{\pars{3}}$ becomes \begin{equation} \left\lbrace\begin{array}{rcl} \ds{\int_{0}^{1}\ln\pars{\Gamma\pars{z}}} & \ds{=} & \ds{\phantom{-\,}\half\,\ln\pars{2\pi}} \\[1mm] \ds{\int_{0}^{1/4}\ln\pars{\Gamma\pars{z}}} & \ds{=} & \ds{\phantom{-\,}{K \over 4\pi} + {1 \over 8}\ln\pars{2\pi} + {9 \over 8}\,\ln\pars{A}} \\[1mm] \ds{\int_{0}^{3/4}\ln\pars{\Gamma\pars{z}}} & \ds{=} & \ds{-\,{K \over 4\pi} + {3 \over 8}\ln\pars{2\pi} + {9 \over 8}\,\ln\pars{A}} \end{array}\right.\tag{6} \end{equation}

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With $\ds{\pars{6}}$, the expression $\ds{\pars{2}}$ is reduced to $\ds{\pars{~\ul{the\ final\ result}~}}$: $$ \color{#f00}{\int_{0}^{\infty}\bracks{% {x - 1 \over \ln^{2}\pars{x}} - {1 \over \ln\pars{x}}} \,{\dd x \over x^{2} + 1}} = \color{#f00}{4\,{K \over \pi}} \approx 1.1662 $$