2

In my post, I had found the values of the couple of integrals, $$ \int_{0}^{\frac{\pi}{4}} x \tan x d x=-\frac{\pi}{8} \ln 2+\frac{G}{2}\tag*{(1)} $$ and $$ \int_{0}^{\frac{\pi}{4}} x \cot x d x=\frac{\pi}{8} \ln 2+\frac{G}{2}\tag*{(2)} $$ I just wonder what happens when I combine these 2 results and consequently find a wonderful result $$ \int_{0}^{\frac{\pi}{2}} \frac{x}{\sin x} d x=2 G. $$

Adding (1) and (2) yields $$ G=\int_{0}^{\frac{\pi}{4}} x \tan x d x+ \int_{0}^{\frac{\pi}{4}} x \cot x d x$$

$$ \begin{aligned} \because x \tan x+x \cot x &=x\left(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right) \\ &=\frac{x}{\cos x \sin x} \\ &=\frac{2 x}{\sin (2 x)} \end{aligned} $$

$$ \therefore \int_{0}^{\frac{\pi}{4}} x \tan x d x+\int_{0}^{\frac{\pi}{4}} x \cot x d x=\int_{0}^{\frac{\pi}{4}} \frac{2 x}{\sin (2 x)} d x $$Letting $x\mapsto \frac{x}{2}$ yields $$ G=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{x}{\sin x} d x $$ Now we can conclude that $$ \int_{0}^{\frac{\pi}{2}} \frac{x}{\sin x} d x=2 G. $$

I am curious whether there are more elegant proofs.

Your comments and alternate solutions are warmly welcome.

Lai
  • 20,421
  • 1
    $$\frac{dx}{\sin x}=d\big(\ln(\tan\frac{x}{2})\big),,\Rightarrow,,I=\int_{0}^{\frac{\pi}{2}} \frac{x}{\sin x} d x=\int_{0}^{\frac{\pi}{2}}xd\big(\ln(\tan\frac{x}{2})\big)$$ $$=x\ln(\tan\frac{x}{2})\Big|0^\frac{\pi}{2}-\int{0}^{\frac{\pi}{2}}\ln(\tan\frac{x}{2})dx=-2\int_{0}^{\frac{\pi}{4}}\ln(\tan t)dt$$ Making the substitution $\tan t =x$ $$I=-2\int_0^1\frac{\ln x}{1+x^2}dx=2G$$ – Svyatoslav Jan 20 '22 at 13:53
  • 1
    A nice proof! Thank you very much! – Lai Jan 20 '22 at 15:42

1 Answers1

1

Inspired by Mr Svyatoslav,

$$ \begin{aligned} \int_{0}^{\frac{\pi}{2}} \frac{x}{\sin x} d x &=4 \int_{0}^{\frac{\pi}{2}} x d\left(\ln \left(\tan \frac{x}{2}\right)\right) \\ &=4\left[x \ln \left(\tan \frac{x}{2}\right)\right]_{0}^{\frac{\pi}{2}}-4 \int_{0}^{\frac{\pi}{2}} \ln \left(\tan \frac{x}{2}\right) d x \\\\ &=-2 \int_{0}^{\frac{\pi}{4}} \ln (\tan x) d x \end{aligned} $$

By my post $$ \int_{0}^{\frac{\pi}{4}} \ln (\tan x) d x=-G,$$

we can conclude that $$ \int_{0}^{\frac{\pi}{2}} \frac{x}{\sin x} d x=2 G $$

Lai
  • 20,421