# An AMM-like integral $\int_0^1\frac{\arctan x}x\ln\frac{(1+x^2)^3}{(1+x)^2}dx$

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.

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

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.

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.

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 $$\log(1+x^2)=\log(1+ix)+\log(1-ix)$$.

• Thank you for the great answer. :) But I prefer a solution without brute force. Nov 21, 2018 at 3:02

@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

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