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I need help on proving $(1)$. $$I=\int_{0}^{\pi\over 4}\arctan{(\cot^2{x})}\mathrm dx={2\pi^2-\ln^2({3+2\sqrt{2})}\over 16}\tag1$$ This is what I have attempted;

Enforcing a sub: $u=\cot^2{x}$ then $du=-2\cot{x}\csc^2{x}dx$

Recall $1+\cot^2{x}=\csc^2{x}$

$$I={1\over2}\int_{1}^{\infty}\arctan{u}\cdot{\mathrm dx\over u^{1/2}+u^{3/2}}$$

Recall $u^3+1=(u+1)(u^2-u+1)$

$$I={1\over2}\int_{1}^{\infty}\arctan{u}\left({A\over u^{1/2}}+{B\over u+1}+{Cu+D\over u^2-u+1}\right)\mathrm du$$

I am stuck at this point.

Can anyone help to prove $(1)$?

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  • $\begingroup$ 1) Where do you get these integrals from? 2) If you let $u=\cot x$ you almost end up with the integral you posted yesterday (I have not seen if that leads all the way). $\endgroup$ – mickep Jan 11 '17 at 7:10
  • $\begingroup$ @mickep do I get a cup of tea if I tell you where I got the idea of this integral from? math.stackexchange.com/questions/1287169/… $\endgroup$ – user339807 Jan 11 '17 at 7:30
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    $\begingroup$ Providing such information may improve your chance of getting an answer. For instance, your link clearly shows the relationship between your integral and the Legendre chi function, which indeed turns out to be a crucial hint toward the computation. $\endgroup$ – Sangchul Lee Jan 11 '17 at 7:45
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We first write $I$ as

$$ I= \frac{\pi^2}{16} + \int_{0}^{\frac{\pi}{4}} \left( \arctan(\cot^2 x) - \arctan(1) \right) \, dx. $$

Now using addition formulas for $\arctan$ and $\cos$, we have

$$ \arctan(\cot^2 x) - \arctan(1) = \arctan\left(\frac{\cot^2 x - 1}{\cot^2 x + 1} \right) = \arctan(\cos 2x). $$

Consequently we have

\begin{align*} I &= \frac{\pi^2}{16} + \int_{0}^{\frac{\pi}{4}} \arctan(\cos 2x) \, dx \\ &= \frac{\pi^2}{16} + \frac{1}{2}\int_{0}^{\frac{\pi}{2}} \arctan(\sin \theta) \, d\theta, \end{align*}

where the last line follows from the substitution $\theta = \frac{\pi}{2} - 2x$. The last integral can be computed in terms of the Legendre chi function $\chi_2$:

$$ \int_{0}^{\frac{\pi}{2}} \arctan(\sin \theta) \, d\theta = 2\chi_2(\sqrt{2}-1). \tag{1} $$

For a proof of $\text{(1)}$, see my previous answer for instance. There are only a handful of known special values of $\chi_2$, but thankfully

$$\chi_2(\sqrt{2}-1) = \frac{\pi^2}{16} - \frac{1}{4}\log^2(\sqrt{2}+1) \tag{2} $$

is one of them. Summarizing, we have

$$ I = \frac{\pi^2}{8} - \frac{1}{4}\log^2(\sqrt{2}+1), $$

which coincides with the proposed answer.


Addendum. The identity $\text{(2)}$ follows by plugging $x = \sqrt{2}-1$ to the identity

$$ \chi_2\left(\frac{1-x}{1+x}\right) + \chi_2(x) = \frac{\pi^2}{8} - \frac{1}{2}\log x \log\left(\frac{1-x}{1+x}\right), $$

which can be easily checked by differentiating both sides.

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    $\begingroup$ Solid, as always. Sometimes I think you are a machine! :) $\endgroup$ – mickep Jan 11 '17 at 7:11
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    $\begingroup$ @mickep, Thank you! I was lucky enough that I quickly recognized the similarity between the proposed answer and the special value $\text{(2)}$. As I already knew the relation $\text{(1)}$, the rest was to manipulate the integral so that it fits into $\text{(1)}$. $\endgroup$ – Sangchul Lee Jan 11 '17 at 7:16
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    $\begingroup$ (+1) Man you didn't give me time to think about an approach. $\endgroup$ – Zaid Alyafeai Jan 11 '17 at 7:18
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    $\begingroup$ @SangchulLee, I did get the expression for $I$ before the relation $(1)$ as in your answer and was thinking how to proceed, but by that time you already posted the answer with an addendum. (+1) for the superb speed! $\endgroup$ – Samrat Mukhopadhyay Jan 11 '17 at 7:20
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    $\begingroup$ what a nice approach! (+1)super 440! $\endgroup$ – Renascence_5. Jan 11 '17 at 8:18
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For the evaluation of $\displaystyle J=\int_0^{\tfrac{\pi}{2}}\arctan(\sin x)dx$

Perform the change of variable $y=\sin x$,

$\displaystyle J=\int_0^1 \dfrac{\arctan x}{\sqrt{1-x^2}}dx$

Perform the change of variable $y=\sqrt{\dfrac{1-x}{1+x}}$,

$\begin{align}\displaystyle J&=2\int_0^1 \dfrac{\arctan\left(\tfrac{1-x^2}{1+x^2}\right)}{1+x^2}dx\\ &=2\int_0^1 \dfrac{\arctan(1)}{1+x^2}dx-2\int_0^1 \dfrac{\arctan(x^2)}{1+x^2}dx\\ &=\dfrac{\pi^2}{8}-2\int_0^1 \dfrac{\arctan(x^2)}{1+x^2}dx\\ \end{align}$

$\begin{align} \displaystyle\int_0^1 \dfrac{\arctan(x^2)}{1+x^2}dx&=\Big[\arctan x\arctan(x^2)\Big]_0^1-\int_0^1 \dfrac{2x\arctan x}{1+x^4}dx\\ &=\dfrac{\pi^2}{16}-\int_0^1 \dfrac{2x\arctan x}{1+x^4}dx\\ \end{align}$

Since,

$\displaystyle \arctan x=\int_0^1 \dfrac{x}{1+t^2x^2}dx$

then,

$\begin{align} \displaystyle K&=\int_0^1 \dfrac{2x\arctan x}{1+x^4}dx\\ \displaystyle &=\int_0^1\int_0^1 \dfrac{2x^2}{(1+t^2x^2)(1+x^4)}dtdx\\ \displaystyle &=\int_0^1\int_0^1 \left(\dfrac{2t^2}{(1+t^4)(1+x^4)}+\dfrac{2x^2}{(1+x^4)(1+t^4)}-\dfrac{2t^2}{(1+t^4)(1+t^2x^2}\right)dtdx\\ &=\displaystyle 4\left(\int_0^1 \dfrac{t^2}{1+t^4}dt\right)\left(\int_0^1 \dfrac{1}{1+x^4}dx\right)-K \end{align}$

Therefore,

$\displaystyle K=2\left(\int_0^1 \dfrac{x^2}{1+x^4}dx\right)\left(\int_0^1 \dfrac{1}{1+x^4}dx\right)$

Since,

$\begin{align}\displaystyle \int_0^1 \dfrac{x^2}{1+x^4}dx&=\left[\dfrac{1}{4\sqrt{2}}\ln\left(\dfrac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}\right)+\dfrac{1}{2\sqrt{2}}\arctan\left(\sqrt{2}x+1\right)+\dfrac{1}{2\sqrt{2}}\arctan\left(\sqrt{2}x-1\right)\right]_0^1\\ &=\dfrac{1}{4\sqrt{2}}\Big(\pi+\ln\left(3-2\sqrt{2}\right)\Big) \end{align}$

and,

$\begin{align}\displaystyle \int_0^1 \dfrac{1}{1+x^4}dx&=\left[\dfrac{1}{4\sqrt{2}}\ln\left(\dfrac{x^2+\sqrt{2}x+1}{x^2-\sqrt{2}x+1}\right)+\dfrac{1}{2\sqrt{2}}\arctan\left(\sqrt{2}x+1\right)+\dfrac{1}{2\sqrt{2}}\arctan\left(\sqrt{2}x-1\right)\right]_0^1\\ &=\dfrac{1}{4\sqrt{2}}\Big(\pi-\ln\left(3-2\sqrt{2}\right)\Big) \end{align}$

Therefore,

$\boxed{K=\displaystyle \dfrac{\pi^2}{16}-\dfrac{1}{16}\Big(\ln\left(3-2\sqrt{2}\right)\Big)^2}$

Therefore,

$\boxed{\displaystyle\int_0^1 \dfrac{\arctan(x^2)}{1+x^2}dx=\dfrac{1}{16}\Big(\ln\left(3-2\sqrt{2}\right)\Big)^2}$

Therefore,

$\boxed{\displaystyle J=\dfrac{\pi^2}{8}-\dfrac{1}{8}\Big(\ln\left(3-2\sqrt{2}\right)\Big)^2}$

Addendum:

$\displaystyle I=\int_0^{\tfrac{\pi}{4}} \arctan\left(\cot^2(x)\right)dx$

Perform the change of variable,

$y=\tan x$,

$\begin{align}I&=\displaystyle \int_0^1 \dfrac{\arctan\left(\tfrac{1}{x^2}\right)}{1+x^2}dx\\ &=\displaystyle \int_0^1 \dfrac{\tfrac{\pi}{2}-\arctan\left(x^2\right)}{1+x^2}dx\\ &=\displaystyle \dfrac{\pi^2}{8}-\int_0^1 \dfrac{\arctan\left(x^2\right)}{1+x^2}dx \end{align}$

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  • $\begingroup$ Oh man, this is an amazing answer, especially since it doesn't rely on any other nontrivial prerequisite theorems (like special values of the Legendre Chi function). Nicely done, sir! (+1) $\endgroup$ – Franklin Pezzuti Dyer Jun 23 '18 at 21:49

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