15
$\begingroup$

In order to compute, in an elementary way,

$\displaystyle \int_0^1 \frac{x \arctan x \log \left( 1-x^2\right)}{1+x^2}dx$

(see Evaluating $\int_0^1 \frac{x \arctan x \log \left( 1-x^2\right)}{1+x^2}dx$)

i need to show, in a simple way, that:

$\displaystyle \int_0^1 \dfrac{\arctan x \log x}{1+x}dx=\dfrac{G\ln 2}{2}-\dfrac{\pi^3}{64}$

$G$ is the Catalan's constant.

$\endgroup$
  • 3
    $\begingroup$ What is your question? What does simple way mean? Do you already know some other way which you are not satisfied with? $\endgroup$ – Yuriy S Jun 28 '16 at 9:59
  • 3
    $\begingroup$ Moreover, why do you think there is an elementary way to evaluate the linked integral? $\endgroup$ – Yuriy S Jun 28 '16 at 10:04
  • 1
    $\begingroup$ "simple way": use of change of variable, double, triple integrals, derivation under integral and such stuff. Anyway, if you have a solution, whatever it's, i'm interrested. I don't know if a simple solution do exist. $\endgroup$ – FDP Jun 28 '16 at 10:14
21
$\begingroup$

I finally get a solution (i swear i didn't know it when i have posted the question)

Define for $x\in [0,1]$ the function $F$:

$\displaystyle F(x)=\int_0^x \dfrac{\ln t}{1+t}dt$

Notice that $F(1)=-\dfrac{\pi^2}{12}$

(use Taylor's development)

and, after performing the change of variable $y=\dfrac{t}{x}$,

$\displaystyle F(x)=\int_0^1 \dfrac{x\ln(xy)}{1+xy}dy$

Since that:

$\Big[F(x)\arctan x\Big]_0^1=-\dfrac{\pi^3}{48}$

then,

$\displaystyle -\dfrac{\pi^3}{48}=\int_0^1 \dfrac{F(x)}{1+x^2}dx+\int_0^1 \dfrac{\arctan x\ln x}{1+x}dx$

$\displaystyle\int_0^1 \dfrac{F(x)}{1+x^2}dx=\int_0^1\int_0^1 \dfrac{x\ln(xy)}{(1+xy)(1+x^2)}dxdy$

$\displaystyle\int_0^1 \dfrac{F(x)}{1+x^2}dx=\int_0^1\int_0^1 \dfrac{x\ln(x)}{(1+xy)(1+x^2)}dxdy+\int_0^1\int_0^1 \dfrac{x\ln(y)}{(1+xy)(1+x^2)}dxdy$

$\displaystyle\int_0^1 \dfrac{F(x)}{1+x^2}dx=\int_0^1\left[\dfrac{\ln x\ln(1+xy)}{1+x^2}\right]_{y=0}^{y=1} dx+ \displaystyle \int_0^1 \left[-\dfrac{\ln y\ln(1+xy)}{1+y^2}+\dfrac{\ln y\ln(1+x^2)}{2(1+y^2)}+\dfrac{y\ln y\arctan x}{1+y^2}\right]_{x=0}^{x=1}dy$

$\displaystyle\int_0^1 \dfrac{F(x)}{1+x^2}dx= \int_0^1 \dfrac{\ln x\ln(1+x)}{1+x^2}dx-\int_0^1\dfrac{\ln y\ln(1+y)}{1+y^2}dy+\dfrac{\ln 2}{2}\int_0^1 \dfrac{\ln y}{1+y^2}dy+ \dfrac{\pi}{4}\times \int_0^1 \dfrac{y\ln y}{1+y^2}dy$

Using Taylor's development,

$\displaystyle \int_0^1 \dfrac{y\ln y}{1+y^2}dy=-\dfrac{\pi^2}{48}$

And it's well known that, $\displaystyle -G=\int_0^1\dfrac{\ln y}{1+y^2}dy$

Therefore,

$\displaystyle\int_0^1 \dfrac{F(x)}{1+x^2}dx=-\dfrac{G\ln 2}{2}-\dfrac{\pi^3}{192}$

And finally,

$\displaystyle \int_0^1 \dfrac{\arctan x \ln x}{1+x}dx=\dfrac{G\ln 2}{2}-\dfrac{\pi^3}{64}$

(I hope there is no mistake, this proof is too wonderful to be true )

NB:

Added, July 2, 2019.

The above computation is the result of "reverse engineering". I was searching for a way to express $\pi^3$ as in integral. If you introduce the function, for $x\in [0;1]$, \begin{align}\displaystyle F(x)&=\int_0^x \dfrac{\ln t}{1+t}dt\\ &=\int_0^1 \dfrac{x\ln(tx)}{1+tx}dt \end{align} Observe that, \begin{align}\frac{\partial F(x)}{\partial x}&=\dfrac{\ln x}{1+x}\\ F(1)&=-\frac{\pi^2}{12} \end{align}

Then, \begin{align}-\frac{\pi^3}{48}&=\Big[F(x)\arctan x\Big]_0^1\\ \end{align} And, \begin{align}\frac{\partial F(x)}{\partial x}\arctan x=\frac{\arctan x\ln x}{1+x}\end{align}

Thus, one can apply integration by parts, \begin{align}\int_0^1 \frac{\arctan x\ln x}{1+x}\,dx&=\int_0^1 \frac{\partial F(x)}{\partial x}\arctan x\,dx\end{align} and so on,

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ If there is no error. It's too wonderful to be true , i can't believe that's true $\endgroup$ – FDP Jun 28 '16 at 13:59
  • $\begingroup$ I am not able to spot any mistake. It is a nice trick to work with two variables, it happens very often that by sticking to single-variable manipulations life gets hard in these cases. $\endgroup$ – Jack D'Aurizio Jun 28 '16 at 14:06
  • $\begingroup$ very elegant approach. $\endgroup$ – Ali Shadhar May 1 '19 at 13:38
7
$\begingroup$

Let we deal with a basic problem first, i.e. the computation of $$ C_{2n+1} = \int_{0}^{1}\frac{x^{2n+1}\log x}{1+x}\,dx = \int_{0}^{+\infty}\frac{t e^{-(2n+2)t}}{1+e^{-t}}\,dt\tag{1}$$ Since $\int_{0}^{+\infty}t e^{-mt}\,dt = \frac{1}{m^2}$, we have: $$ -C_{2n+1} = \frac{1}{(2n+2)^2}-\frac{1}{(2n+3)^2}+\frac{1}{(2n+4)^2}-\ldots=\frac{\psi'(n+1)-\psi'\left(n+\frac{3}{2}\right)}{4}\tag{2}$$ and: $$ I=\int_{0}^{1}\frac{\arctan(x)\log(x)}{1+x}\,dx = -\sum_{n\geq 0}\frac{(-1)^n C_{2n+1}}{2n+1}=-\sum_{m\geq 0}\sum_{n\geq 0}\frac{(-1)^{n+m}}{(2n+1)(2n+m+2)^2}\tag{3}$$ By reindexing the last double series, $$ I = -\sum_{s=0}^{+\infty}\sum_{p=0}^{s}\frac{(-1)^s}{(2p+1)(p+s+2)^2}=-\sum_{p=0}^{+\infty}\sum_{s\geq p}\frac{(-1)^s}{(p+s+2)^2(2p+1)}\tag{4}$$ hence, in terms of the Hurwitz zeta function: $$ I = -\sum_{p\geq 0}\frac{(-1)^p}{4(p+1)}\left(\zeta\left(2,p+1\right)-\zeta\left(2,p+\frac{3}{2}\right)\right)\tag{5}$$ or, by using the inverse Laplace transform: $$ I = -\int_{0}^{+\infty}\frac{s e^{s/2}\log(1+e^{-s})}{4(1+e^{s/2})}\,ds =-\int_{0}^{+\infty}\frac{s e^s \log(1+e^{-2s})}{1+e^s}\,ds\tag{6}$$ where the last integral is a bit more manageable than the initial one (we made the arctangent function disappear). The constants $K,\log 2$ and $$ \sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^3}=\frac{\pi^3}{32} \tag{7}$$ (see here for the last identity) should simply appear by integration by parts.


With a suitable change of variable and differentiation under the integral sign, we may probably also exploit the integral remainder term in the second Binet's formula for $\log\Gamma$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Very Nice :) (+1) $\endgroup$ – Behrouz Maleki Jun 28 '16 at 14:10
  • $\begingroup$ Nice but too clever for me $\endgroup$ – FDP Jun 28 '16 at 14:38
2
$\begingroup$

Hint:

set $x=e^{-y}$ we have \begin{align} & \int_{0}^{1}{\frac{{{\tan }^{-1}}x\,\,\ln x}{1+x}}\,dx=\int_{0}^{\infty }{\,\frac{-y\,{{e}^{-y}}{{\tan }^{-1}}({{e}^{-y}})\,}{1+{{e}^{-y}}}}\,dy \\ \\ & {-{e}^{-y}}{{\tan }^{-1}}({{e}^{-y}})=-{e}^{-y}\sum\limits_{n=1}^{\infty }{\frac{{{(-1)}^{n+1}}}{2n-1}{{e}^{-(2n-1)y}}}=\sum\limits_{n=1}^{\infty }{\frac{{{(-1)}^{n}}}{2n-1}{{e}^{-2n\,y}}} \\ \\ & \frac{1}{1+{{e}^{-y}}}=\sum\limits_{n=0}^{\infty }{{{(-1)}^{n}}{{e}^{-ny}}} \\ \end{align}

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ ok, but what now? $\endgroup$ – tired Jun 28 '16 at 10:27
  • $\begingroup$ so complex but it is heuristic. $\endgroup$ – Behrouz Maleki Jun 28 '16 at 10:30
  • 1
    $\begingroup$ factor $ln x$ isn't the problem. $\displaystyle \int_0^1 x^n\ln x dx=-\dfrac{1}{(1+n)^2}$ $\endgroup$ – FDP Jun 28 '16 at 10:53
  • 3
    $\begingroup$ Do you really know if this hint actually leads to an elementary way of computing the integral? Judging by your comment it is not clear. $\endgroup$ – Najib Idrissi Jun 28 '16 at 12:25
2
$\begingroup$

This is a long solution but I hope you find it useful.

First lets consider the integral: \begin{align*} I&=\int_0^1\frac{\ln x\arctan x}{x(1+x)}\ dx\\ &=\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\int_0^1\frac{x^{2n}\ln x}{1+x}\ dx\\ &=\frac12\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\frac{\partial}{\partial{n}}\int_0^1\frac{x^{2n}}{1+x}\ dx\\ &=\frac12\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\frac{\partial}{\partial{n}}\left(H_n-H_{2n}+\ln2\right)\\ &=\frac12\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\left(2H_{2n}^{(2)}-H_n^{(2)}-\zeta(2)\right)\\ &=\frac12\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\left(2H_{2n}^{(2)}-H_n^{(2)}\right)-\frac{\pi^3}{48}\tag{1} \end{align*}

on the other hand, \begin{align*} I=\int_0^1\frac{\ln x\arctan x}{x(1+x)}\ dx=\int_0^1\frac{\ln x\arctan x}{x}\ dx-\int_0^1\frac{\ln x\arctan x}{1+x}\ dx\tag{2} \end{align*} where \begin{align*} \int_0^1\frac{\ln x\arctan x}{x}\ dx&=\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\int_0^1x^{2n}\ln x\ dx=-\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^3}=-\frac{\pi^3}{32} \end{align*}

we can conclude from $(1)$ and $(2)$ that \begin{align*} \int_0^1\frac{\ln x\arctan x}{1+x}\ dx&=-\frac{\pi^3}{96}-\frac12\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\left(2H_{2n}^{(2)}-H_n^{(2)}\right)\\ &=-\frac{\pi^3}{96}-\frac12\left(2S_1-S_2\right)\tag{3} \end{align*} \begin{align} S_1&=\sum_{n=0}^\infty\frac{(-1)^nH_{2n}^{(2)}}{2n+1}\\ &=\sum_{n=0}^\infty\frac{(-1)^nH_{2n+1}^{(2)}}{2n+1}-\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^3}\\ &\boxed{=\Im\sum_{n=1}^\infty\frac{i^nH_n^{(2)}}{n}-\frac{\pi^3}{32}=S_1} \end{align} \begin{align} S_2&=\sum_{n=0}^\infty\frac{(-1)^nH_{n}^{(2)}}{2n+1}\\ &=\sum_{n=0}^\infty(-1)^nH_n^{(2)}\int_0^1x^{2n}\ dx\\ &=\int_0^1\sum_{n=0}^\infty H_n^{(2)}(-x^2)^n\\ &=\int_0^1\frac{\operatorname{Li}_2(-x^2)}{1+x^2}\ dx \quad \text{ apply IBP}\\ &=-\frac{\pi^3}{48}+2\int_0^1\frac{\arctan x\ln(1+x^2)}{x}\ dx\tag{#}\\ &=-\frac{\pi^3}{48}-4\sum_{n=0}^\infty\frac{(-1)^nH_{2n}}{2n+1}\int_0^1x^{2n}\ dx\\ &=-\frac{\pi^3}{48}-4\sum_{n=0}^\infty\frac{(-1)^nH_{2n}}{(2n+1)^2}\\ &=-\frac{\pi^3}{48}-4\sum_{n=0}^\infty\frac{(-1)^nH_{2n+1}}{(2n+1)^2}+4\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^3}\\ &=-\frac{\pi^3}{48}-4\sum_{n=0}^\infty\frac{(-1)^nH_{2n+1}}{(2n+1)^2}+4\times\frac{\pi^3}{32}\\ &\boxed{=-4\Im\sum_{n=1}^\infty\frac{i^nH_n}{n^2}+\frac{5\pi^3}{48}=S_2} \end{align} note that in line $\text{(#)}$, we used $\ \displaystyle\arctan x\ln(1+x^2)=-2\sum_{n=0}^{\infty}\frac{(-1)^n H_{2n}} {2n+1}x^{2n+1}\ $ (see here).

Plugging $S_1$ and $S_2$ in $(3)$, we get $$\int_0^1\frac{\arctan x\ln x}{1+x}\ dx=\frac{7\pi^3}{96}-\Im\left(\sum_{n=1}^\infty\frac{i^nH_n^{(2)}}{n}+2\sum_{n=1}^\infty\frac{i^nH_n}{n^2}\right)$$

using the generating functions: $$\sum_{n=1}^\infty\frac{x^nH_n^{(2)}}{n}=\operatorname{Li}_3(x)+2\operatorname{Li}_3(1-x)-\ln(1-x)\operatorname{Li}_2(1-x)-\zeta(2)\ln(1-x)-2\zeta(3)$$

$$\sum_{n=1}^\infty\frac{x^nH_n}{n^2}=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\frac12\ln^2x\ln(1-x)+\zeta(3)$$ then \begin{align} \sum_{n=1}^\infty\frac{x^nH_n^{(2)}}{n}+2\sum_{n=1}^\infty\frac{x^nH_n}{n^2}&=3\operatorname{Li}_3(x)+\ln(1-x)\{\operatorname{Li}_2(1-x)+\ln x\ln(1-x)-\zeta(2)\}\\ &=3\operatorname{Li}_3(x)-\ln(1-x)\operatorname{Li}_2(x) \end{align} where in the last line, we used the reflection identity. taking $x=i$ , we get \begin{align} \Im\left(\sum_{n=1}^\infty\frac{i^nH_n^{(2)}}{n}+2\sum_{n=1}^\infty\frac{i^nH_n}{n^2}\right)&=\Im\left(3\operatorname{Li}_3(i)-\ln(1-i)\operatorname{Li}_2(i)\right)\\ &=\frac{17\pi^3}{192}-\frac12G\ln2 \end{align} which follows \begin{align} \int_0^1\frac{\arctan x\ln x}{1+x}\ dx&=\frac{7\pi^3}{96}-\left(\frac{17\pi^3}{192}-\frac12G\ln2\right)\\ &=\frac12G\ln2-\frac{\pi^3}{64} \end{align}

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Any thoughts on how to sum those Euler sums? $\endgroup$ – omegadot Jun 15 '19 at 6:26
  • $\begingroup$ @omegadot the first sum is easy, as for the second one , I haven't tried it yet. $\endgroup$ – Ali Shadhar Jun 15 '19 at 9:32
  • $\begingroup$ @omegadot you can see a full solution now. I did little manipulation with these two sums. they can be evaluated separately but I chose a short cut. hope you like it. $\endgroup$ – Ali Shadhar Jul 2 '19 at 1:22
  • 2
    $\begingroup$ Very nice. The magic key here to evaluating these Euler sums is by using appropriate generating functions evaluated at $x = i$. $\endgroup$ – omegadot Jul 2 '19 at 2:13
  • $\begingroup$ @omegadot yes i would say thats the key but i am sure there is a much shorter hidden way to manipulate with these two sums. i will investigate into that. $\endgroup$ – Ali Shadhar Jul 2 '19 at 2:25
2
$\begingroup$

Different approach:

start with applying integration by parts

$$I=\int_0^1\frac{\tan^{-1}(x)\ln(x)}{1+x}dx\\=\left|(\operatorname{Li}_2(-x)+\ln(x)\ln(1+x))\tan^{-1}(x)\right|_0^1-\int_0^1\frac{\operatorname{Li}_2(-x)+\ln(x)\ln(1+x)}{1+x^2}dx$$

$$=-\frac{\pi^3}{48}-\int_0^1\frac{\operatorname{Li}_2(-x)}{1+x^2}dx-\color{blue}{\int_0^1\frac{\ln(x)\ln(1+x)}{1+x^2}dx}\tag1$$


From $$\operatorname{Li}_2(x)=-\int_0^1\frac{x\ln(y)}{1-xy}dy$$

it follows that

$$\int_0^1\frac{\operatorname{Li}_2(-x)}{1+x^2}dx=\int_0^1\frac1{1+x^2}\left(\int_0^1\frac{x\ln(y)}{1+xy}dy\right)dx$$

$$=\int_0^1\ln(y)\left(\int_0^1\frac{x}{(1+x^2)(1+yx)}dx\right)dy$$

$$=\int_0^1\ln(y)\left(\frac{\pi}{4}\frac{y}{1+y^2}-\frac{\ln(1+y)}{1+y^2}+\frac{\ln(2)}{2(1+y^2)}\right)dy$$

$$=-\frac{\pi^3}{192}-\color{blue}{\int_0^1\frac{\ln(y)\ln(1+y)}{1+y^2}dy}-\frac12\ln(2)\ G\tag2$$

By plugging $(2)$ in $(1)$, the blue integral magically cancels out and we get $I=\frac12G\ln2-\frac{\pi^3}{64}$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ @omegadot i think you will like this answer :) $\endgroup$ – Ali Shadhar Nov 20 '19 at 23:36
  • $\begingroup$ I do indeed. Now if only one could find a value for the blue integral. It pops up in a number of problems like this which I have seen, before magically disappearing again. $\endgroup$ – omegadot Dec 30 '19 at 6:09
  • $\begingroup$ @omegadot I think the blue integral is manageable we can just evaluate $\int_0^1 \frac{\operatorname{Li}_2(-x)}{1+x^2}dx$ by converting the boundaries from $(0,1)$ to $(0,\infty)$ and we follow the same steps above. $\endgroup$ – Ali Shadhar Dec 30 '19 at 6:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.