I was able to calculate: $$\int_0^\infty\arctan\left(e^{-x}\right)\,dx=G$$ $$\int_0^\infty\arctan^2\left(e^{-x}\right)\,dx=\frac\pi2\,G-\frac78\zeta(3)$$ $G$ is the Catalan constant. In both cases Maple is able to find corresponding indefinite integral, and the above results can be obtained by taking limits. They also can be done using identities $$\arctan\left(e^{-x}\right)=\Im\left(\log\left(1+ie^{-x}\right)\right)\color{gray}{,\quad x\in\mathbb{R}}$$ $$\arctan^2\left(e^{-x}\right)=\frac14\log^2\left(1+e^{-2x}\right)-\Re\left(\log^2\left(1+ie^{-x}\right)\right)\color{gray}{,\quad x\in\mathbb{R}}$$ I'm also interested in this value: $$\int_0^\infty\arctan\left(e^{-x}\right)\,\arctan\left(e^{-2x}\right)\,dx$$ Could you help me with it? Maple cannot do it.
Need help with $\int_0^\infty\arctan\left(e^{-x}\right)\,\arctan\left(e^{-2x}\right)\,dx$
1.2k Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail AtThere are 6 best solutions below
On
Too long for a comment :
As far as definite integrals are concerned, there are no currently known methods of actually proving that they do not possess a closed form. $($ In other words, there are no equivalents of Liouville's theorem or the Risch algorithm for definite integrals $).$ So what I'm about to write
does not constitute “proof” of anything, but serves merely as an intuitive explanation for why
it is unlikely that the integral you posted to have a “meaningful” closed form. Notice that the
other two integrands possess an inverse, i.e., $$y~=~\arctan^n(e^{-x})~\iff~x~=~-\ln\tan\sqrt[n]y~=~\ln\cos\sqrt[n]y~-~\ln\sin\sqrt[n]y~.$$ At the same time, remember the well-known relation between the definite integral of a function, and that of its inverse:
$\qquad\qquad\qquad\quad$
Applying it to the former two cases, it yields something manageable, following a substitution of the form $y=t^n.$ The latter integrand, however, does not exhibit such auspicious traits, since its inverse cannot be expressed in terms of any known functions, be they either special or elementary.
On
Using the series for arctan, it is $$\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\frac{(-1)^{i+j}}{(2i+1)(4j+2)(2i+4j+3)}$$
On
We'll have use for the following trigonometric identity, which may be verified by the well-known arctangent addition law:
$$\small{\arctan{\left(x^{2}\right)}=\arctan{\left(\frac{x}{\sqrt{2}-x}\right)}-\arctan{\left(\frac{x}{\sqrt{2}+x}\right)}};~~~\small{\left|x\right|<\sqrt{2}}.$$
We'll also need the definition of the arctangent function in terms of complex logarithms:
$$\arctan{\left(z\right)}=\frac{\ln{\left(\frac{1+iz}{1-iz}\right)}}{2i}.$$
Then,
$$\begin{align} I &=\int_{0}^{\infty}\arctan{\left(e^{-y}\right)}\arctan{\left(e^{-2y}\right)}\,\mathrm{d}y\\ &=\int_{0}^{1}\frac{\arctan{\left(x\right)}\arctan{\left(x^{2}\right)}}{x}\,\mathrm{d}x;~~~\small{\left[e^{-y}=x\right]}\\ &=\int_{0}^{1}\frac{\arctan{\left(x\right)}\left[\arctan{\left(\frac{x}{\sqrt{2}-x}\right)}-\arctan{\left(\frac{x}{\sqrt{2}+x}\right)}\right]}{x}\,\mathrm{d}x\\ &=\int_{0}^{1}\frac{\arctan{\left(x\right)}\left[\ln{\left(1-e^{-\frac{i\pi}{4}}x\right)}-\ln{\left(1-e^{\frac{i\pi}{4}}x\right)}\right]}{2ix}\,\mathrm{d}x\\ &~~~~~-\int_{0}^{1}\frac{\arctan{\left(x\right)}\left[\ln{\left(1+e^{\frac{i\pi}{4}}x\right)}-\ln{\left(1+e^{-\frac{i\pi}{4}}x\right)}\right]}{2ix}\,\mathrm{d}x\\ &=\small{\int_{0}^{1}\frac{\left[\ln{\left(1+ix\right)}-\ln{\left(1-ix\right)}\right]\left[\ln{\left(1+e^{\frac{i\pi}{4}}x\right)}-\ln{\left(1+e^{-\frac{i\pi}{4}}x\right)}\right]}{4x}\,\mathrm{d}x}\\ &~~~~~\small{-\int_{0}^{1}\frac{\left[\ln{\left(1+ix\right)}-\ln{\left(1-ix\right)}\right]\left[\ln{\left(1+e^{\frac{3i\pi}{4}}x\right)}-\ln{\left(1+e^{-\frac{3i\pi}{4}}x\right)}\right]}{4x}\,\mathrm{d}x}\\ &=\small{\int_{0}^{1}\frac{\Re{\left[\ln{\left(1+ix\right)}\ln{\left(1+e^{\frac{i\pi}{4}}x\right)}-\ln{\left(1-ix\right)}\ln{\left(1+e^{\frac{i\pi}{4}}x\right)}\right]}}{2x}\,\mathrm{d}x}\\ &~~~~~\small{-\int_{0}^{1}\frac{\Re{\left[\ln{\left(1+ix\right)}\ln{\left(1+e^{\frac{3i\pi}{4}}x\right)}-\ln{\left(1-ix\right)}\ln{\left(1+e^{\frac{3i\pi}{4}}x\right)}\right]}}{2x}\,\mathrm{d}x}.\\ \end{align}$$
Any integral of the form $\int\frac{\ln{\left(ax+b\right)}\ln{\left(cx+d\right)}}{px+q}\,\mathrm{d}x$ can be systematically reduced to trilogarithms, dilogarithms, and elementary functions. Thus, we recognize that $I$ can be decomposed into a sum of such integrals, and so in principle we are done except for the tedium.
On
Let $I$ denotes your integral, $C$ denotes Catalan's constant. $$ \begin{aligned} I&=\int_{0}^{1} \frac{\arctan(x)\arctan(x^2)}{x} \text{d}x\\ &=\frac{1}{2} \int_{0}^{1} \frac{\arctan(x)\arctan(\sqrt{x} )}{x}\text{d}x\\ &= \frac{\pi C}{8}-\frac{1}{4} \int_{0}^{1} \frac{\operatorname{Ti}_2(x)}{\sqrt{x}(1+x) } \text{d}x\\ &=\frac{3\pi C}{8}-\frac{1}{8} \int_{0}^{\infty} \frac{\operatorname{Ti}_2(x)}{\sqrt{x}(1+x) } \text{d}x\\ &=\frac{3\pi C}{8}-\frac{1}{8}\underbrace{ \int_{-\infty}^{\infty} \frac{\operatorname{Ti}_2(x^2)}{1+x^2 } \text{d}x}_{J}\\ \end{aligned} $$ Notice that $\operatorname{Ti}_2(x^2)=\Im\operatorname{Li}_2(ix^2)$. And $\operatorname{Li}_2(iz^2) =2 \left ( \operatorname{Li}_2(e^{\pi i/4}z) +\operatorname{Li}_2(-e^{\pi i/4}z) \right ).$ So $$ J=2\Im\left ( \int_{-\infty}^{\infty} \frac{\operatorname{Li}_2(e^{\pi i/4}z) +\operatorname{Li}_2(-e^{\pi i/4}z)}{1+z^2}\text{d}z \right ). $$ Consider the functions $$ f(z)=\frac{\operatorname{Li}_2(e^{\pi i/4}z)}{1+z^2} , f^{\prime}(z) =\frac{\operatorname{Li}_2(-e^{\pi i/4}z)}{1+z^2}. $$ $f(z),f'(z)$ have two branch points in the lower half plane and upper half plane, respectively. Now we construct two very large semicircles in upper and lower half plane such that $f(z),f'(z)$ are meromorphic in the interiors. By Cauchy residue theorem, $$ J=4\pi\Im\left ( \operatorname{Li}_2(ie^{\pi i/4}) \right ). $$ And $$I=\frac{3\pi C}{8}-\frac{\pi}{2} \Im\left ( \operatorname{Li}_2(ie^{\pi i/4}) \right ).$$ Therefore $I$ is evaluated as $$I=\frac{\pi C}{2} -\frac{\pi\left ( \psi^{(1)}\left ( \frac{1}{8} \right ) +\psi^{(1)}\left ( \frac{3}{8} \right ) -\psi^{(1)}\left ( \frac{5}{8} \right ) -\psi^{(1)}\left ( \frac{7}{8} \right )\right ) }{128\sqrt{2} }$$ where $\psi^{(1)}(z)=\sum_{n=0}^\infty\frac{1}{(n+z)^2}$ is the derivative of digamma function $\psi^{(0)}(z)$.
On
This is a solution within real analysis by hand that avoids polylogarithms $$I=\int_0^\infty\arctan(e^{-x})\arctan(e^{-2x})dx=\int_0^1\frac{\arctan x\arctan x^2 }{x}dx$$ Integrate by parts with $u=\arctan x\arctan x^2$ and $dv=\frac{dx}{x}$ $$I=-\int_0^1\left(\frac{2x\ln x\arctan x}{x^4+1}+\frac{\ln x\arctan x^2}{x^2+1}\right)dx=-2J-K$$ Start with J. We invert the integral with $u=\frac{1}{x}$, then add two versions of J. Then, isolate $J$ $$J=\int_0^1\frac{x\ln x\arctan x}{x^4+1}dx=\int_1^\infty\frac{x\ln x \left(\arctan x-\frac{\pi}{2}\right)}{x^4+1}dx$$ $$2J=-\frac{\pi G}{8}+\int_0^\infty\frac{x\ln x\arctan x}{x^4+1}dx$$ $$J=\frac{-\pi G}{16}+\frac{1}{2}\int_0^\infty\frac{x\ln x \arctan x}{x^4+1}dx$$ This gives an improper integral. Substitute an integral representation of the inverse tangent function, and apply Fubini's theorem. $$=\frac{-\pi G}{16}+\frac{1}{2}\int_0^\infty\frac{x\ln x}{x^4+1}\left(\int_0^1\frac{x}{x^2y^2+1}dy\right)dx$$ $$=\frac{-\pi G}{16}+\frac{1}{2}\int_0^1\int_0^\infty\frac{x^2\ln x}{(x^2y^2+1)(x^4+1)}dx\space dy$$ Evaluate the inside integral separately with partial fractions $$\int_0^\infty\frac{x^2\ln x}{(x^2y^2+1)(x^4+1)}dx=\int_0^\infty\left(\frac{y^2}{y^4+1}\frac{1}{x^4+1}+\frac{1}{y^4+1}\frac{x^2}{x^4+1}-\frac{y^2}{y^4+1}\frac{1}{x^2y^2+1}\right)\ln x \space dx$$$$=-\frac{\pi^2}{8\sqrt2}\frac{y^2}{y^4+1}+\frac{\pi^2}{8\sqrt2}\frac{1}{y^4+1}-\frac{y}{y^4+1}\int_0^\infty\frac{\ln x}{x^2y^2+1}y\space dx$$ Substitute $x→\frac{x}{y}$ $$=-\frac{\pi^2}{8\sqrt2}\frac{y^2}{y^4+1}+\frac{\pi^2}{8\sqrt2}\frac{1}{y^4+1}-\frac{y}{y^4+1}\int_0^\infty\frac{\ln x-\ln y}{x^2+1}\space dx$$$$=-\frac{\pi^2}{8\sqrt2}\frac{y^2}{y^4+1}+\frac{\pi^2}{8\sqrt2}\frac{1}{y^4+1}+\frac{\pi}{2}\frac{y\ln{y}}{y^4+1}$$ Plug back into the equation for $J$ $$J=\frac{-\pi G}{16}+\frac{1}{2}\int_0^1\left(-\frac{\pi^2}{8\sqrt2}\frac{y^2}{y^4+1}+\frac{\pi^2}{8\sqrt2}\frac{1}{y^4+1}+\frac{\pi}{2}\frac{y\ln{y}}{y^4+1}\right)dy$$$$=\frac{-\pi G}{16}-\frac{\pi^2}{16\sqrt2}\left(\frac{\pi}{4\sqrt2}+\frac{\ln(\sqrt2-1)}{2\sqrt2}\right)+\frac{\pi^2}{16\sqrt2}\left( \frac{\pi}{4\sqrt2}+\frac{\ln(1+\sqrt2)}{2\sqrt2} \right)+\frac{\pi}{4}\left(\frac{-G}{4}\right)$$$$=\frac{-\pi G}{8}+\frac{\pi^2\ln(1+\sqrt2)}{32}$$ Moving on to K. We can apply the same method. Invert, add 2 versions, isolate $K$ $$K=\int_0^1\frac{\ln x \arctan x^2}{x^2+1}dx=\int_1^\infty\frac{\ln x (\arctan x^2-\frac{\pi}{2})}{x^2+1}dx$$ $$2K=\frac{-\pi G}{2}+\int_0^\infty\frac{\ln x\arctan x^2}{x^2+1}dx$$$$K=\frac{-\pi G}{4}+\frac{1}{2}\int_0^\infty\frac{\ln x\arctan x^2}{x^2+1}dx$$ Insert an integral representation for $\arctan{x^2}$ and apply Fubini's theorem$$=\frac{-\pi G}{4}+\frac{1}{2}\int_0^\infty\frac{\ln x}{x^2+1}\left(\int_0^1\frac{2x^2y}{x^4y^4+1}dy\right)dx$$$$=\frac{-\pi G}{4}+\int_0^1\int_0^\infty\frac{x^2y\ln{x}}{(x^4y^4+1)(x^2+1)}dx\space dy$$ Evaluate the inside integral separately. Apply partial fractions, substitute $x→\frac{x}{y}$, distribute, and integrate term-by-term $$\int_0^\infty\frac{x^2y\ln{x}}{(x^4y^4+1)(x^2+1)}dx=\int_0^\infty\left(\frac{y^5}{y^4+1}\frac{x^2}{x^4y^4+1}+\frac{y}{y^4+1}\frac{1}{x^4y^4+1}-\frac{y}{y^4+1}\frac{1}{x^2+1}\right)\ln x\space dx$$$$=\int_0^\infty\left(\frac{y^2}{y^4+1}\frac{x^2y^2}{x^4y^4+1}+\frac{1}{y^4+1}\frac{1}{x^4y^4+1}\right)y\ln x\space dx$$$$=\int_0^\infty\left(\frac{y^2}{y^4+1}\frac{x^2}{x^4+1}+\frac{1}{y^4+1}\frac{1}{x^4+1}\right)(\ln x-\ln y)dx=\frac{\pi^2}{8\sqrt2}\frac{y^2-1}{y^4+1}-\frac{\pi}{2\sqrt2}\frac{y^2+1}{y^4+1}\ln y$$ Plug it back into the equation for $K$ $$K=\frac{-\pi G}{4}+\int_0^1\left(\frac{\pi^2}{8\sqrt2}\frac{y^2-1}{y^4+1}-\frac{\pi}{2\sqrt2}\frac{y^2+1}{y^4+1}\ln y\right)dy$$$$=\frac{-\pi G}{4}+\frac{\pi^2}{8\sqrt2}\frac{\ln(\sqrt2-1)}{\sqrt2}-\frac{\pi}{2\sqrt2}\int_0^1\frac{y^2+1}{y^4+1}\ln y\space dy$$ Evaluate the integral on the right-hand side separately. Apply geometric series, integrate the series term-by-term, and apply the definition of the trigamma function. $$\int_0^1\frac{y^2+1}{y^4+1}\ln y\space dy=\int_0^1\frac{1+y^2-y^4-y^6}{1-y^8}\ln y \space dy$$$$=\int_0^1(1+y^2-y^4-y^6)\ln y\sum_{n=0}^\infty y^{8n}dy$$$$=\sum_{n=0}^\infty\int_0^1(y^{8n}+y^{8n+2}-y^{8n+4}-y^{8n+6})\ln y\space dy$$$$=\sum_{n=0}^\infty\left(\frac{1}{(8n+7)^2}+\frac{1}{(8n+5)^2}-\frac{1}{(8n+3)^2}-\frac{1}{(8n+1)^2}\right)$$$$=\frac{\psi\left(\frac{7}{8}\right)+\psi\left(\frac{5}{8}\right)-\psi\left(\frac{3}{8}\right)-\psi\left(\frac{1}{8}\right)}{64}$$ Plug it back into the equation for $K$ $$K=\frac{-\pi G}{4}+\frac{\pi^2\ln(\sqrt2-1)}{16}-\frac{\pi}{128\sqrt2}\left({\psi\left(\frac{7}{8}\right)+\psi\left(\frac{5}{8}\right)-\psi\left(\frac{3}{8}\right)-\psi\left(\frac{1}{8}\right)}\right)$$ Plug $J$ and $K$ back into the equation for $I$ $$I=-2J-K$$$$=-2\left(\frac{-\pi G}{8}+\frac{\pi^2\ln(1+\sqrt2)}{32}\right)-\left(\frac{-\pi G}{4}+\frac{\pi^2\ln(\sqrt2-1)}{16}-\frac{\pi}{128\sqrt2}\left({\psi\left(\frac{7}{8}\right)+\psi\left(\frac{5}{8}\right)-\psi\left(\frac{3}{8}\right)-\psi\left(\frac{1}{8}\right)}\right)\right)$$$$=\frac{\pi G}{2}+\frac{\pi}{128\sqrt2}\left({\psi\left(\frac{7}{8}\right)+\psi\left(\frac{5}{8}\right)-\psi\left(\frac{3}{8}\right)-\psi\left(\frac{1}{8}\right)}\right)$$ Final answer: $$I=\int_0^\infty\arctan(e^{-x})\arctan(e^{-2x})dx=\frac{\pi G}{2}+\frac{\pi}{128\sqrt2}\left({\psi\left(\frac{7}{8}\right)+\psi\left(\frac{5}{8}\right)-\psi\left(\frac{3}{8}\right)-\psi\left(\frac{1}{8}\right)}\right)$$ We also have $$J=\int_0^1\frac{x\ln x\arctan x}{x^4+1}dx=\frac{-\pi G}{8}+\frac{\pi^2}{32}\ln(1+\sqrt2)$$ $$K=\int_0^1\frac{\ln x\arctan x^2}{1+x^2}dx=\frac{-\pi G}{4}+\frac{\pi^2\ln(\sqrt2-1)}{16}-\frac{\pi}{128\sqrt2}\left({\psi\left(\frac{7}{8}\right)+\psi\left(\frac{5}{8}\right)-\psi\left(\frac{3}{8}\right)-\psi\left(\frac{1}{8}\right)}\right)$$
This is a fascinating integral! Following generally the same approach as shown in David H's answer, with help from Mathematica and manual simplification using known di- and trilogarithm identities, I was able to establish the following: $$\int_0^x\arctan\left(e^{-z}\right)\,\arctan\left(e^{-2z}\right)\,dz=x\operatorname{arccot}\left(e^x\right)\operatorname{arccot}\left(e^{2x}\right)\\ +\frac{\pi^2}{32}\left(\vphantom{\Large|}\!\ln\left(1+e^{4x}\right)-10\ln\left(e^{2x}+e^x\sqrt2+1\right)\right)\\ +\frac\pi2\left(\vphantom{\Large|}\!\operatorname{arccot}\left(1+e^x\sqrt2\right)+\arctan\left(e^x\right)\right)\cdot\ln\left(e^{2x}+e^x\sqrt2+1\right)\\ +\frac i4\left\{\left(\vphantom{\Large|}2\operatorname{arccot}\left(e^x\right)+\operatorname{arccot}\left(e^{2x}\right)\right)\cdot\\ \left[\operatorname{Li}_2\left(\tfrac{(-1)^{3/4}+e^x}{-i+e^x}\right)-\operatorname{Li}_2\left(\tfrac{-(-1)^{1/4}+e^x}{i+e^x}\right) +\operatorname{Li}_2\left(\tfrac{i-(-1)^{1/4}e^x}{i+e^x}\right)-\operatorname{Li}_2\left(\tfrac{1-(-1)^{1/4}e^x}{1+ie^x}\right)\right]\\ +\left(\vphantom{\Large|}\!\operatorname{arccot}\left(e^{2x}\right)+2\operatorname{arccot}\left(1+e^x\sqrt2\right)-2\operatorname{arccot}\left(e^x\right)\right)\cdot\\ \left[\operatorname{Li}_2\left(\tfrac{-(-1)^{1/4}+e^x}{-i+e^x}\right)-\operatorname{Li}_2\left(\tfrac{(-1)^{3/4}+e^x}{i+e^x}\right)+\operatorname{Li}_2\left(\tfrac{-1+(-1)^{1/4}e^x}{-1+ie^x}\right)-\operatorname{Li}_2\left(\tfrac{-i+(-1)^{1/4}e^x}{-i+e^x}\right)\right]\right\}\\ +\frac i2\left\{\operatorname{arccot}\left(1+e^x\sqrt2\right)\cdot\\ \left[\operatorname{Li}_2\left(\tfrac{-(-1)^{3/4}+e^x}{i+e^x}\right)-\operatorname{Li}_2\left(\tfrac{(-1)^{1/4}+e^x}{-i+e^x}\right)+\operatorname{Li}_2\left(\tfrac{i+(-1)^{1/4}e^x}{i-e^x}\right)-\operatorname{Li}_2\left(\tfrac{i+(-1)^{3/4}e^x}{i+e^x}\right)\right]+\left(\vphantom{\Large|}\!\operatorname{arccot}\left(e^x\right)-\operatorname{arccot}\left(1+e^x\sqrt2\right)\right)\cdot\left[\vphantom{\Large|}\operatorname{Li}_2\left(1-(-1)^{1/4}e^x\right)-\operatorname{Li}_2\left(1+(-1)^{1/4}e^x\right)\\ +\operatorname{Li}_2\left(1-(-1)^{3/4}e^x\right)-\operatorname{Li}_2\left(1+(-1)^{3/4}e^x\right)\right]\right\}\\ +\frac14\left[\operatorname{Li}_3\left(\tfrac{-(-1)^{1/4}+e^x}{-i+e^x}\right)-\operatorname{Li}_3\left(\tfrac{-(-1)^{1/4}+e^x}{i+e^x}\right)+\operatorname{Li}_3\left(\tfrac{(-1)^{1/4}+e^x}{-i+e^x}\right)-\operatorname{Li}_3\left(\tfrac{(-1)^{1/4}+e^x}{i+e^x}\right)\\ -\operatorname{Li}_3\left(\tfrac{-(-1)^{3/4}+e^x}{-i+e^x}\right)+\operatorname{Li}_3\left(\tfrac{-(-1)^{3/4}+e^x}{i+e^x}\right) -\operatorname{Li}_3\left(\tfrac{(-1)^{3/4}+e^x}{-i+e^x}\right)+\operatorname{Li}_3\left(\tfrac{(-1)^{3/4}+e^x}{i+e^x}\right)\\ +\operatorname{Li}_3\left(\tfrac{i-(-1)^{1/4}e^x}{i+e^x}\right)-\operatorname{Li}_3\left(\tfrac{-1+(-1)^{1/4}e^x}{-1+ie^x}\right)-\operatorname{Li}_3\left(\tfrac{-i+(-1)^{1/4}e^x}{-i+e^x}\right)-\operatorname{Li}_3\left(-\tfrac{i+(-1)^{1/4}e^x}{-i+e^x}\right)\\ +\operatorname{Li}_3\left(\tfrac{i+(-1)^{1/4}e^x}{i+e^x}\right)+\operatorname{Li}_3\left(\tfrac{-i+(-1)^{3/4}e^x}{-i+e^x}\right)+\operatorname{Li}_3\left(-\tfrac{i+(-1)^{3/4}e^x}{-i+e^x}\right) -\operatorname{Li}_3\left(\tfrac{i+(-1)^{3/4}e^x}{i+e^x}\right)\right]\\ +\frac\pi{64}\left[\left(2+\sqrt2\right)\cdot\left(16G+\pi^2\sqrt2\right)-\sqrt2\,\psi^{(1)}\!\left(\tfrac18\right)\right]$$ (here is the corresponding Mathematica expression)
If I'm not mistaken, it must hold for all real $x$, and might hold for some complex $x$, but in general there are some branch cuts that need to be dealt with (I was not yet able to analyze them completely). A possible proof consists of taking a derivative, a (tedious) simplification and computing a limit to establish the constant of integration. I relied on Mathematica for some steps, followed by high-precision numerical validation.
So, the final answer is