Hellow I am trying to prove this result. $$ I:=\int_0^\infty (1-x\cot^{-1} x)dx=\frac{\pi}{4}. $$ The indefinite integral exists for this integral. The function $\cot^{-1} x$ is the arc-cotangent function, not the multiplicative inverse. Note, we can not just break the integral up into two pieces because we will have problems with divergence.

Using the relation $$ x\cot^{-1} x=x \tan^{-1} \frac{1}{x},\to \quad I=\int_0^\infty \left(1-x\tan^{-1} \frac{1}{x}\right)dx $$ may be of help to some but didn't help me. I am not sure if I am missing a clever substitution, perhaps integration by parts will work, I obtained using this method $$ I=(x-x^2\cot^{-1} x)\big|^\infty_0-\int_0^\infty \frac{x^2}{x^2+1}dx+\int_0^\infty x\cot^{-1} x \, dx $$ but this is clearly a problem since we have divergence issue now.

The indefinite integral is given by $$ \int (1-x\cot^{-1} x ) dx =\frac{1}{2}\left(x+\tan^{-1} x-x^2 \cot^{-1} x\right) $$ but I am unable to prove this result. Thanks.

  • $\begingroup$ @BenjaminDickman I am not sure why you deleted your previous comment. It wasn't very helpful at all but it may help others. See Random Variable solution below. Thanks $\endgroup$ – Jeff Faraci May 19 '14 at 1:29
  • $\begingroup$ (Offtopic) Did you look at my comment on this: math.stackexchange.com/questions/713836/… $\endgroup$ – Pranav Arora May 19 '14 at 7:13

Integrating by parts (as you've done),

$$\begin{align}\int x \cot^{-1}(x) \ dx &= x^{2} \cot^{-1} (x) - \int x\cot^{-1}(x) \ dx + \int \frac{x^{2}}{1+x^{2}} \ dx\\ \implies \int x \cot^{-1}(x) \ dx &= \frac{1}{2} \Big(x^{2} \cot^{-1} (x) + \int \ dx - \int \frac{1}{1+x^{2}} \ dx\Big)\\ &= \frac{1}{2} \Big(x^{2} \cot^{-1} (x) + x - \tan^{-1}(x)\Big).\end{align}$$


$$\begin{align}\int \left( 1- x \cot^{-1}(x) \right) \ dx &= x - \frac{1}{2} \Big(x^{2} \cot^{-1} (x) + x - \tan^{-1}(x)\Big)\\ &= \frac{1}{2} \left(x + \tan^{-1}(x) - x^{2}\cot^{-1}{x} \right).\end{align}$$


And just to add,

$$ \lim_{x \to \infty} \left(x + \tan^{-1}(x) - x^{2}\cot^{-1}(x) \right) = \lim_{x \to \infty} \left[x + \tan^{-1}(x) - x^{2} \left(\frac{1}{x} + \mathcal{O}(x^{-3}) \right) \right] = \frac{\pi}{2}.$$


$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{I\equiv\int_{0}^{\infty}\bracks{1 - x\ {\rm arccot}\pars{x}}\,\dd x ={\pi \over 4}:\ {\large ?}}$

With $\ds{0 < \epsilon < \Lambda}$: $$ \int_{\epsilon}^{\Lambda}\bracks{1 - x\ {\rm arccot}\pars{x}}\,\dd x =\Lambda - \epsilon -\color{#c00000}{\int_{\epsilon}^{\Lambda}x\ {\rm arccot}\pars{x}\,\dd x} $$

\begin{align} &\color{#c00000}{\int_{\epsilon}^{\Lambda}x\ {\rm arccot}\pars{x}\,\dd x} =\int_{\epsilon}^{\Lambda}x\arctan\pars{1 \over x}\,\dd x =-\int_{1/\epsilon}^{1/\Lambda}{\arctan\pars{x} \over x^{3}}\,\dd x \\[3mm]&=\half\,\Lambda^{2}\arctan\pars{1 \over \Lambda} -\half\,\epsilon^{2}\arctan\pars{1 \over \epsilon} -\int_{1/\epsilon}^{1/\Lambda}{1 \over 2x^{2}}\,{1 \over x^{2} + 1}\,\dd x \\[3mm]&=\half\,\Lambda^{2}\arctan\pars{1 \over \Lambda} -\half\,\epsilon^{2}\arctan\pars{1 \over \epsilon} -\half\int_{1/\epsilon}^{1/\Lambda}\pars{{1 \over x^{2}} - {1 \over x^{2} + 1}} \,\dd x \\[3mm]&=\half\,\Lambda^{2}\arctan\pars{1 \over \Lambda} -\half\,\epsilon^{2}\arctan\pars{1 \over \epsilon} + \half\,\Lambda - \half\,\epsilon + \half\,\arctan\pars{1 \over \Lambda} -\half\,\arctan\pars{1 \over \epsilon} \end{align}

Then \begin{align} &\int_{\epsilon}^{\Lambda}\bracks{1 - x\ {\rm arccot}\pars{x}}\,\dd x =\ \overbrace{\half\,\Lambda\bracks{1 - \Lambda\arctan\pars{\Lambda} - {1 \over \Lambda}\arctan\pars{1 \over \Lambda}}} ^{\ds{\to\ 0\quad\mbox{when}\quad\Lambda\ \to\ \infty}} \\[3mm]&+\ \overbrace{\half\,\epsilon\bracks{\epsilon\arctan\pars{1 \over \epsilon} - 1}} ^{\ds{\to\ 0\quad\mbox{when}\quad\epsilon\ \to\ 0^{+}}}\ +\ \half\ \overbrace{\arctan\pars{1 \over \epsilon}} ^{\ds{\to\ {\pi \over 2}\ \mbox{when}\ \epsilon\ \to\ 0^{+}}} \end{align}

$$\color{#00f}{\large% \int_{0}^{\infty}\bracks{1 - x\ {\rm arccot}\pars{x}}\,\dd x={\pi \over 4}} $$

  • 1
    $\begingroup$ You're like... the king of LaTeX ✌.ʕʘ‿ʘʔ.✌ $\endgroup$ – Lucas Zanella May 19 '14 at 4:45
  • 4
    $\begingroup$ His LaTex is good. What makes Felix is a king is that he solves All integrals I post!!!:) +1 (and he also has the ability to solve many other problems as I always see). +1 always for creative original solutions. thanks Felix $\endgroup$ – Jeff Faraci May 19 '14 at 4:52
  • $\begingroup$ Thanks @LucasZanella and $\tt\mbox{@Integrals}$ for your comments. $\endgroup$ – Felix Marin May 19 '14 at 5:06

Let $y=\cot^{-1}x$, then $x=\cot y\;\Rightarrow\;dx=d(\cot y)$. The integral turns out to be $$ \int(1-y\cot y)\ d(\cot y)=\int\ d(\cot y)-\int y\cot y\ d(\cot y).$$ The second integral in the RHS can be solved using IBP by taking $u=y$ and $dv=\cot y\ d(\cot y)$, we obtain $du=dy$ and $v=\dfrac12\cot^2 y$. Hence $$ \int y\cot y\ d(\cot y)=\frac12y\cot^2 y-\frac12\int\cot^2 y\ dy, $$ where $$ \int\cot^2 y\ dy=\int(\csc^2 y-1)\ dy=-\cot y-y+C. $$ Therefore $$ \begin{align} \int(1-y\cot y)\ d(\cot y)&=\cot y-\frac12y\cot^2 y-\frac12\cot y-\frac12y+C\\ &=\frac12(\cot y-y\cot^2 y-y)+C\\ \int (1-x\cot^{-1} x)\ dx&=\frac12\left(x-x^2\cot^{-1}x-\cot^{-1}x\right)+C. \end{align} $$ Actually, the result of integral is as the same result as yours Jeff. To make sure, using the identity $\tan^{-1}x+\cot^{-1}x=\dfrac\pi2$ for $x>0$ yields $$ \begin{align} \int (1-x\cot^{-1} x)\ dx&=\frac12\left(x-x^2\cot^{-1}x\right)-\frac12\cot^{-1}x+C\\ &=\frac12\left(x-x^2\cot^{-1}x\right)-\frac12\cot^{-1}x+\frac\pi4+K\\ &=\frac12\left(x-x^2\cot^{-1}x\right)+\frac12\left(\frac\pi2-\cot^{-1}x\right)+K\\ &=\frac12\left(x-x^2\cot^{-1}x\right)+\frac12\tan^{-1}x+K\\ &=\frac12\left(x+\tan^{-1}x-x^2\cot^{-1}x\right)+K.\qquad\blacksquare \end{align} $$


$\int(1-x\operatorname{arccot}(x))dx=x-\int x\operatorname{arccot}(x)dx$


$=x-x^{2}\operatorname{arccot}(x)-\frac{x}{2}\ln(1+x^{2})+\int x\operatorname{arccot}(x)dx+\frac{1}{2}\int\ln(1+x^{2})dx$

$=x-x^{2}\operatorname{arccot}(x)-\frac{x}{2}\ln(1+x^{2})+\int x\operatorname{arccot}(x)dx+\frac{1}{2}(x\ln(1+x^{2})-2\int\frac{x^{2}}{x^{2}+1}dx)$

$x-x^{2}\operatorname{arccot}(x)+\int x\operatorname{arccot}(x)dx-\int1dx+\int\frac{1}{x^{2}+1}dx$





  • $\begingroup$ Thanks +1, perhaps you can work on formatting your answers in LaTex more properly. I probably would have chose this as the answer however it was a bit troublesome to read. +1 though THanks. $\endgroup$ – Jeff Faraci May 19 '14 at 1:28
  • $\begingroup$ You're welcome. I apologize for the difficulty reading. $\endgroup$ – user71352 May 19 '14 at 1:29
  • $\begingroup$ No problem. Thanks for the solution. $\endgroup$ – Jeff Faraci May 19 '14 at 1:29

Inverting $x\to1/x$ gives

$$\int_0^\infty(1-x\cot^{-1}x)dx=\int_0^\infty{x-\arctan x\over x^3}dx$$

Integration by parts with $u=x-\arctan x$ and $dv=dx/x^3$ turns this into

$${x-\arctan x\over2x^2}\Big|_0^\infty+{1\over2}\int_0^\infty{1\over1+x^2}dx$$

The final integral easily gives the desired $\pi/4$, so it remains to check that

$$\lim_{x\to0}{x-\arctan x\over2x^2}=\lim_{x\to\infty}{x-\arctan x\over2x^2}=0$$

But these limits are easy L'Hopital computations:

$${(x-\arctan x)'\over(2x^2)'}={1-\displaystyle{1\over1+x^2}\over4x}={x\over4(1+x^2)}$$


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