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How can we evaluate $$I=\int_0^1\frac{\arctan x}x\ln\frac{(1+x^2)^3}{(1+x)^2}dx=0?$$

I tried substitution $x=\frac{1-t}{1+t}$ and got $$I=\int_0^1\frac{2 \ln \frac{2 (t^2+1)^3}{(t+1)^4} \arctan \frac{t-1}{t+1}}{t^2-1}dt\\ =\int_0^1\frac{2 \ln \frac{2 (t^2+1)^3}{(t+1)^4} (\arctan t-\frac\pi4)}{t^2-1}dt$$ I'm able to evaluate $$\int_0^1\frac{\ln \frac{2 (t^2+1)^3}{(t+1)^4}}{t^2-1}dt$$ But I have no idea where to start with the rest one.

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  • $\begingroup$ Just wondering: Why do you want to calculate this (by hand) $\endgroup$
    – klirk
    Nov 20, 2018 at 20:38
  • $\begingroup$ @klirk Just an interest. $\endgroup$ Nov 21, 2018 at 0:18
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    $\begingroup$ $$\int_0^1\frac{\ln \frac{2 (t^2+1)^3}{(t+1)^4}}{t^2-1}dt=\frac{\pi^2}{48}$$ $\endgroup$
    – user178256
    Nov 21, 2018 at 11:00

4 Answers 4

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A solution by Cornel Ioan Valean. The problem is similar to the problem AMM $12054$. Using the well-known result in $4.535.1$ from Table of Integrals, Series and Products by I.S. Gradshteyn and I.M. Ryzhik: $$\int_0^1 \frac{\arctan(y x)}{1+y^2x}\textrm{d}x=\frac{1}{2y^2}\arctan(y)\log(1+y^2)$$ We have: $$\frac{1}{2}\int_0^1\frac{\arctan(y)\log(1+y^2)}{y}dy=\int_0^1\left(\int_0^1 \frac{y\arctan(y x)}{1+y^2x}\textrm{d}x\right)\textrm{d}y$$ $$\overset{yx=t}{=}\int_0^1\left(\int_0^y \frac{\arctan(t)}{1+y t}\textrm{d}t\right)\textrm{d}y=\int_0^1\left(\int_t^1 \frac{\arctan(t)}{1+y t}\textrm{d}y\right)\textrm{d}t$$ $$=\int_0^1\frac{\arctan(t)\log\left(\frac{1+t}{1+t^2}\right)}{t} \textrm{d}t\overset{t=y}=\int_0^1\frac{\arctan(y)\log\left(\frac{1+y}{1+y^2}\right)}{y} \textrm{d}y$$ And the result is proved.

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A (new) solution by Cornel Ioan Valean

It's straightforward to show by symmetry means that

\begin{equation*} \int_0^1 \left(\int_0^1 \frac{a x}{(1+a^2 x^2) (1+a^2 x y)} \textrm{d}x \right)\textrm{d}y=\int_0^1\frac{a x}{1+a^2 x^2}\textrm{d}x\int_0^1 \frac{1}{1+a^2 y^2}\textrm{d}y \end{equation*} \begin{equation*} =\frac{\arctan(a)\log(1+a^2)}{2a^2}. \end{equation*}

The exact flow is described in the book (Almost) Impossible Integrals, Sums, and Series, page $162$, where the only difference is that we inject a parameter $a$, that is we use $ax$ instead of $x$ and $ay$ instead of $y$.

If we multiply the opposite sides of the result above by $a$ and then integrate from $a=0$ to $a=1$, we get \begin{equation*} \frac{1}{2}\int_0^1 \frac{\arctan(a)\log(1+a^2)}{a}\textrm{d}a=\int_0^1\left(\int_0^1 \left(\int_0^1 \frac{a^2 x}{(1+a^2 x^2) (1+a^2 x y)} \textrm{d}x \right)\textrm{d}y\right)\textrm{d}a \end{equation*} \begin{equation*} =\int_0^1\left(\int_0^1 \left(\int_0^1 \frac{a^2 x}{(1+a^2 x^2) (1+a^2 x y)} \textrm{d}y \right)\textrm{d}x\right)\textrm{d}a=\int_0^1 \left(\int_0^1 \frac{\log(1+a^2 x)}{1+a^2 x^2}\textrm{d}x\right)\textrm{d}a \end{equation*} \begin{equation*} \overset{a x\mapsto x}{=}\int_0^1\left(\int_0^a \frac{\log(1+a x)}{a(1+x^2)}\textrm{d}x\right)\textrm{d}a=\int_0^1\frac{1}{1+x^2}\left(\int_x^1 \frac{\log(1+a x)}{a}\textrm{d}a\right)\textrm{d}x \end{equation*} \begin{equation*} \overset{x a\mapsto a}{=}\int_0^1\frac{1}{1+x^2}\left(\int_{x^2}^x \frac{\log(1+a)}{a}\textrm{d}a\right)\textrm{d}x \end{equation*} \begin{equation*} =2\int_0^1 \frac{\arctan(x)\log(1+x^2)}{x}\textrm{d}x-\int_0^1 \frac{\arctan(x)\log(1+x)}{x}\textrm{d}x, \end{equation*} and the result follows.

Q.E.D.

A spectacular generalization of the main integral

\begin{equation*} 3 \int_0^x \frac{\arctan(t)\log(1+t^2)}{t} \textrm{d}t=2\int_0^x \frac{\arctan(t) \log (1+x t)}{t} \textrm{d}t \end{equation*}

The proof follows the strategy above where we integrate from $a=0$ to $a=r$, and $r$ is any real number.


Using the same strategy as in the main integral, we may show that

\begin{equation*} 3 \int_0^1 \frac{\arctan(x)\operatorname{Li}_2(x) }{x} \textrm{d}x+\int_0^1 \frac{\arctan(x)\operatorname{Li}_2(-x)}{x} \textrm{d}x-\int_0^1 \frac{\arctan(x)\operatorname{Li}_2\left(-x^2\right)}{x} \textrm{d}x \end{equation*} \begin{equation*} =3 \zeta(2)G+\frac{45 }{16}\zeta (4)-\frac{1}{256}\psi ^{(3)}\left(\frac{1}{4}\right), \end{equation*}

without calculating each integral separately. This problem was prepared for Romanian Mathematical Magazine.

More spectacular results

If we use the result \begin{equation*} \int_0^x\frac{\arctan(t)\log(1+t^2)}{t} \textrm{d}t-2 \int_0^1 \frac{\arctan(xt) \log (1-t)}{t}\textrm{d}t=2\sum_{n=1}^{\infty}(-1)^{n-1} \frac{x^{2n-1}}{(2n-1)^3}, \end{equation*} which is found in the book, (Almost) Impossible Integrals, Sums, and Series, or its extended version which exploits $\displaystyle \operatorname{Ti}_3(x)=\int_0^x\frac{\operatorname{Ti}_2(y)}{y}\textrm{d}y$, \begin{equation*} \int_0^x\frac{\arctan(t)\log(1+t^2)}{t} \textrm{d}t-2 \int_0^1 \frac{\arctan(xt) \log (1-t)}{t}\textrm{d}t=2 \operatorname{Ti}_3(x), \end{equation*} together with the generalization above (included in the paper), we obtain the amazing results

\begin{equation*} \int_0^x \frac{\arctan(t) \log (1+x t)}{t} \textrm{d}t-3 \int_0^1 \frac{\arctan(xt) \log (1-t)}{t}\textrm{d}t=3\sum_{n=1}^{\infty}(-1)^{n-1} \frac{x^{2n-1}}{(2n-1)^3}, \end{equation*}

and if we use the extended version to $\displaystyle \operatorname{Ti}_3(x)$, then we have

\begin{equation*} \int_0^x \frac{\arctan(t) \log (1+x t)}{t} \textrm{d}t-3 \int_0^1 \frac{\arctan(xt) \log (1-t)}{t}\textrm{d}t=3\operatorname{Ti}_3(x). \end{equation*}

For example, setting $x=1$ above, we obtain the special case

\begin{equation*} \int_0^1 \frac{\arctan(t) \log (1+t)}{t} \textrm{d}t-3 \int_0^1 \frac{\arctan(t) \log (1-t)}{t}\textrm{d}t=\frac{3}{32}\pi^3. \end{equation*}

Let's go a bit further and notice that if we exploit the inverse relation of $\operatorname{Ti}_3(x)$, we obtain that

\begin{equation*} \int_0^x\frac{\arctan(t)\log(1+t^2)}{t} \textrm{d}t+\int_0^{1/x}\frac{\arctan(t)\log(1+t^2)}{t} \textrm{d}t \end{equation*} \begin{equation*} -2 \int_0^1 \frac{\arctan(xt) \log (1-t)}{t}\textrm{d}t-2 \int_0^1 \frac{\arctan(t/x) \log (1-t)}{t}\textrm{d}t \end{equation*} \begin{equation*} =\operatorname{sgn}(x)\left(\frac{\pi^3}{8}+\frac{\pi}{2}\log^2(|x|)\right), \end{equation*}

and

\begin{equation*} \int_0^x \frac{\arctan(t) \log (1+x t)}{t} \textrm{d}t+\int_0^{1/x} \frac{\arctan(t) \log (1+t/x)}{t} \textrm{d}t \end{equation*} \begin{equation*} -3 \int_0^1 \frac{\arctan(xt) \log (1-t)}{t}\textrm{d}t-3 \int_0^1 \frac{\arctan(t/x) \log (1-t)}{t}\textrm{d}t \end{equation*} \begin{equation*} =\operatorname{sgn}(x)3\left(\frac{\pi^3}{16}+\frac{\pi}{4}\log^2(|x|)\right). \end{equation*}

Let me now present a new fancy representation of $\pi$ with the results above

\begin{equation*} \large \pi \end{equation*} \begin{equation*} =2\int_0^e\frac{\arctan(t)\log(1+t^2)}{t} \textrm{d}t+2\int_0^{1/e}\frac{\arctan(t)\log(1+t^2)}{t} \textrm{d}t \end{equation*} \begin{equation*} -4 \int_0^1 \frac{\arctan(et) \log (1-t)}{t}\textrm{d}t-4 \int_0^1 \frac{\arctan(t/e) \log (1-t)}{t}\textrm{d}t \end{equation*} \begin{equation*} -4\int_0^1\frac{\arctan(t)\log(1+t^2)}{t} \textrm{d}t+8 \int_0^1 \frac{\arctan(t) \log (1-t)}{t}\textrm{d}t. \end{equation*}

And another new fancy representation of $\pi$

\begin{equation*} \large \pi \end{equation*} \begin{equation*} =\frac{4}{3}\int_0^e \frac{\arctan(t) \log (1+et)}{t} \textrm{d}t+\frac{4}{3}\int_0^{1/e} \frac{\arctan(t) \log (1+t/e)}{t} \textrm{d}t \end{equation*} \begin{equation*} -4 \int_0^1 \frac{\arctan(et) \log (1-t)}{t}\textrm{d}t-4 \int_0^1 \frac{\arctan(t/e) \log (1-t)}{t}\textrm{d}t \end{equation*} \begin{equation*} -\frac{8}{3}\int_0^1 \frac{\arctan(t) \log (1+t)}{t} \textrm{d}t+8 \int_0^1 \frac{\arctan(t) \log (1-t)}{t}\textrm{d}t. \end{equation*}

A note: All the results may be found in the new preprint, A symmetry-related treatment of two fascinating sums of integrals by C.I. Valean.

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Through the dilogarithm/trilogarithm machinery it can be shown that

$$ \int_{0}^{1}\frac{\log(1+i x)\log(1+x)}{x}\,dx=\\\frac{\pi K}{2}-\frac{9i\pi^3}{64}+3iK\log(2)-\frac{3\pi i}{16}\log^2(2)+\frac{5\pi^2}{32}\log(2)-\frac{\log^3(2)}{8}-\frac{69}{16}\zeta(3)+6\,\text{Li}_3\left(\tfrac{1+i}{2}\right) $$

$$ \int_{0}^{1}\frac{\log^2(1+i x)}{x}\,dx=\\ -\frac{\pi K}{2}-\frac{3i\pi^3}{64}+iK\log(2)-\frac{\pi i}{16}\log^2(2)+\frac{5\pi^2}{96}\log(2)-\frac{\log^3(2)}{24}-\frac{3}{16}\zeta(3)+2\,\text{Li}_3\left(\tfrac{1+i}{2}\right) $$

$$ \int_{0}^{1}\frac{\log(1+ix)\log(1-ix)}{x}\,dx= \frac{\pi K}{2}-\frac{27}{32}\zeta(3)$$ hence the claim follows by $\arctan x=\text{Im}\,\log(1+ix)$ and $\log(1+x^2)=\log(1+ix)+\log(1-ix)$.

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  • $\begingroup$ Thank you for the great answer. :) But I prefer a solution without brute force. $\endgroup$ Nov 21, 2018 at 3:02
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@Kemono Chen proved here

$$\int_0^y\frac{\ln(1+yx)}{1+x^2}dx=\frac12 \tan^{-1}(y)\ln(1+y^2)$$

Divide both sides by $y$ then integrate between $0$ and $1$ we get

$$\color{red}{\frac12\mathcal{I}}=\frac12\int_0^1\frac{\tan^{-1}(y)\ln(1+y^2)}{y}dy=\int_0^1\int_0^y\frac{\ln(1+yx)}{y(1+x^2)}dxdy$$

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

$$\overset{IBP}{=}\int_0^1\tan^{-1}(x)\left(\frac{2\ln(1+x^2)}{x}-\frac{\ln(1+x)}{x}\right)dx\\=\color{red}{2\mathcal{I}}-\int_0^1\frac{\tan^{-1}(x)\ln(1+x)}{x}dx$$

which can be written as

$$\int_0^1\tan^{-1}(x)\ln\left(\frac{(1+x^2)^3}{(1+x)^2}\right)dx=0$$

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