8
$\begingroup$

Can anyone help to provide a proof for $(1)$? Pleases! Thank you.

$$\int_{0}^{\pi\over 2}\arctan(\tan^8{(\pi^2{x}}))\mathrm dx={5\over 4}\tag1$$

Enforcing $u=\tan^8{(\pi^2{x})}$ then $du={8\over \pi^2}\tan^7{(\pi^2{x})}\sec^2{(\pi^2{x})}dx$

Recall: $1+\tan^2{x}=\sec^2{x}$

$du={8\over \pi^2}\tan^7{(\pi^2{x})}+{8\over \pi^2}\tan^8{(\pi^2{x})}dx$

$${\pi^2\over 8}\int_{0}^{k}\arctan{u}\cdot{\mathrm du\over u+u^{7/8}}$$

$k=\arctan{\left(\tan^8{\left(\pi^3\over 2\right)}\right)}$

This is where I got so far. Can't go any further.

Extra note

I think this is the correct version

$$\lim_{n\to \infty}\int_{0}^{\pi\over 2}\arctan(\tan^{2n}{(\pi^2{x}}))\mathrm dx={5\over 4}\tag2$$

  • 0
    Please take care of the limits of integration. When $x $ goes from $ 0 \to \frac {\pi}{2} $, $u $ goes from $0 \to \arctan (\tan^8 (\pi^3/2)) $. – 2017-01-10
  • 4
    I think this is wrong. Numerically I get 1.249999992164761... instead of 5/4. Please can anybody confirm this? – 2017-01-10
  • 2
    @Nemo confirmed: http://www.wolframalpha.com/input/?i=NIntegrate%5BArcTan%5BTan%5E8%5BPi%5E2+x%5D%5D,%7Bx,0,Pi%2F2%7D,WorkingPrecision-%3E32%5D Because there is nothing special about $\tan(\pi^3/2)$ such a simple result would be a real suprise – 2017-01-10
  • 10
    Here is an explanation why this integral is so close to $5/4$ but not equal to $5/4$. Let's rewrite it as $$ I=\frac{1}{\pi^2}\int_{0}^{\pi^3\over 2}\arctan(\tan^8{({x}})) dx. $$ It is known that $$ \frac{1}{\pi^2}\int_{0}^{5\pi}\arctan(\tan^8{({x}})) dx=5/4. $$ Then due to $\pi^2\approx 10$ $$ I=\frac54+\frac{1}{\pi^2}\int_{5\pi}^{\pi^3\over 2}\arctan(\tan^8{({x}})) dx\approx\frac54+\frac{1}{\pi^2}\int_0^{{\pi^3\over2}-5\pi }x^8 dx=5/4+\frac{({\pi^3/2}-5\pi )^9}{9\pi^2}=1.24999999285... $$ which is pretty close to the numerical value 1.249999992164761... – 2017-01-10

0 Answers 0