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

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.

• It's correct to at least 200 digits (according to Maple). Jul 18 '16 at 17:05
• It's simply true :D Jul 18 '16 at 17:12
• I just posted yet a fifth way forward. Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark Jul 19 '16 at 15:12
• I am astonished how beautiful all answers are ! Thank you everyone for the hard work and for convincing me in the conjecture :) (OP). Jul 19 '16 at 20:49
• You're welcome! It was my pleasure. Jul 19 '16 at 22:41

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

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

• (+1) It is interesting to point out that since the imaginary parts have to cancel out, that also gives a proof of $\sum_{n\geq 0}\frac{1}{(2n+1)^2}=\frac{\pi^2}{8}.$ Jul 18 '16 at 18:04
• ...or that $\sum_{n\geq0}\frac{(-1)^n}{2n+1}=\frac{\pi}{4}$ :-p Jul 18 '16 at 18:05
• @tired: I hope you don't mind, I added your approach to my Community Wiki answer in the historical thread math.stackexchange.com/questions/8337/… Jul 18 '16 at 18:46
• @JackD'Aurizio i'm fine with that...glad that i also made it into this thread somehow...^^ Jul 18 '16 at 18:54

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.

• By the way, you can also go directly into the first integral without any symmetry considerations...but your approach is definitly cleaner (no nasty imaginary parts which have to cancel) Jul 18 '16 at 17:39
• I put in an approach, have a look if you are interested :) Jul 18 '16 at 18:00
• Jul 18 '16 at 18:29
• @venus, M.N.C.E answer shows that the two integrals have very similar structure Jul 18 '16 at 20:17
• Jack, this is very efficient. I just posted a solution that avoids the complex plane and the use of special functions. Instead, it relies on simple juggling of integrals including an application of Feynman's Trick. The result is a well-known integral representation of $G$. -Mark Jul 19 '16 at 15:11

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!

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.

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!

• @tired Well, after reading all of the great answers already posted, I decided to try another way forward for this one. Jul 19 '16 at 15:05
• Simply a fantastic answer. The beauty in this one is that it only requires first year calc knowledge (along with Fubini's theorem or the like to justify the integral switch and the Catalan's Constant definition) Jul 19 '16 at 16:59
• I have to admit that your answer along with @RandomVariable's are better than mine if not the best of all answers here. Both of you are deserved more than (+1) Jul 20 '16 at 1:55

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: $$\psi(-2,x)=\zeta'(-1,x)-\frac{x^2}{2}+\frac{x}{2}-\frac{1}{12}$$

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.

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 $s=2$.


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}

$\ds{\Psi}$ is the Digamma Function and we used its well known integral representation $\ds{\pars{~\gamma\ \mbox{is the}\ Euler\mbox{-}Mascheroni\ Constant~}}$ $$\Psi\pars{z} = -\gamma + \int_{0}^{1}{1 - t^{z - 1} \over 1 - t}\,\dd t\,, \qquad\Re\pars{z} > 0$$
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}
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, $$\left\lbrace\begin{array}{\rcl} \ds{\int_{0}^{1}\ln\pars{\Gamma\pars{z}}} & \ds{=} & \ds{\half\,\ln\pars{2\pi}\ \mbox{because}\ \Gamma\pars{1} = \,\mathrm{G}\pars{2} = 1.} \\[3mm] \ds{\int_{0}^{1/4}\ln\pars{\Gamma\pars{z}}} & \ds{=} & \ds{{3 \over 32} + {1 \over 8}\,\ln\pars{2\pi} + {1 \over 4}\,\ln\pars{\Gamma\pars{1 \over 4}} - \ln\pars{\,\mathrm{G}\pars{5 \over 4}}} \\[1mm] & \ds{=} & \ds{{3 \over 32} + {1 \over 8}\,\ln\pars{2\pi} - {3 \over 4}\,\ln\pars{\Gamma\pars{1 \over 4}} - \ln\pars{\,\mathrm{G}\pars{1 \over 4}}} \\[3mm] \ds{\int_{0}^{3/4}\ln\pars{\Gamma\pars{z}}} & \ds{=} & \ds{{3 \over 32} + {3 \over 8}\,\ln\pars{2\pi} + {3 \over 4}\,\ln\pars{\Gamma\pars{3 \over 4}} - \ln\pars{\,\mathrm{G}\pars{7 \over 4}}} \\[1mm] & \ds{=} & \ds{{3 \over 32} + {3 \over 8}\,\ln\pars{2\pi} - {1 \over 4}\,\ln\pars{\Gamma\pars{3 \over 4}} - \ln\pars{\,\mathrm{G}\pars{3 \over 4}}} \end{array}\right.\tag{3}$$ 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 $$\left\lbrace\begin{array}{rcl} \ds{\int_{0}^{1}\ln\pars{\Gamma\pars{z}}} & \ds{=} & \ds{\phantom{-\,}\half\,\ln\pars{2\pi}} \\[1mm] \ds{\int_{0}^{1/4}\ln\pars{\Gamma\pars{z}}} & \ds{=} & \ds{\phantom{-\,}{K \over 4\pi} + {1 \over 8}\ln\pars{2\pi} + {9 \over 8}\,\ln\pars{A}} \\[1mm] \ds{\int_{0}^{3/4}\ln\pars{\Gamma\pars{z}}} & \ds{=} & \ds{-\,{K \over 4\pi} + {3 \over 8}\ln\pars{2\pi} + {9 \over 8}\,\ln\pars{A}} \end{array}\right.\tag{6}$$
With $\ds{\pars{6}}$, the expression $\ds{\pars{2}}$ is reduced to $\ds{\pars{~\ul{the\ final\ result}~}}$: $$\color{#f00}{\int_{0}^{\infty}\bracks{% {x - 1 \over \ln^{2}\pars{x}} - {1 \over \ln\pars{x}}} \,{\dd x \over x^{2} + 1}} = \color{#f00}{4\,{K \over \pi}} \approx 1.1662$$

• The OP's integral looked familiar, until I recalled a similar integral in your answer, namely $$\int_0^1\left(\frac{x-1}{\ln^2 x}-\frac{x}{\ln x}\right)\frac{\mathrm{d}x}{x^2+1}=\frac{2G}{\pi}+\frac{\ln 2}6+\frac{\ln\pi}2-6\ln A$$ with $A$ as the Glaisher–Kinkelin constant. Jul 25 '19 at 16:46