Calcular la siguiente integral definida

Question

¿Cómo puedo enfoque y calcular la siguiente integral? Podría polylogarithm funciones y los números complejos se pueden evitar?

$$ \int_0^{\frac{\pi}{2}} \frac{x}{\tan x} dx$$

Answer 1

Integrar por partes con $u=x$ $\displaystyle dv=\frac{dx}{\tan x}$ $$ \int^{\pi/2}_0 \frac{x}{\tan x} dx = x \log (\sin x) \biggr|^{\pi/2}_0 - \int^{\pi/2}_0 \log (\sin x) dx=- \int^{\pi/2}_0 \log (\sin x) dx. $$

Deje $x=\pi/2 - u$ a $\displaystyle I= \int^{\pi/2}_0 \log(\sin x) dx $ conseguir $\displaystyle I=\int^{\pi/2}_0 \log(\cos x) dx.$ Agregar da $$2I =\int^{\pi/2}_0 \log( \sin x \cos x) dx = \int^{\pi/2}_0 \log(\sin 2x) dx - \frac{\pi \log 2}{2}.$$

Ahora por $u=2x$ tenemos $$\int^{\pi/2}_0 \log (\sin 2x) dx = \int^{\pi}_0 \log(\sin u) \frac{du}{2} = I$$

por lo $$\int^{\pi/2}_0 \frac{x}{\tan x} dx = \frac{\pi \log 2}{2}.$$

Answer 2

$$\int_0^{\frac {\pi}2} \frac x{\tan x}\, dx=\left[x\log(\sin (x))\right]_0^{\frac{\pi}2}-\int_0^{\frac {\pi}2} \log(\sin (x))\,dx$$ (el primer término es $0$ en el límite, vamos a evaluar la integral) $$\int_0^{\frac {\pi}2} \log(\sin (x))\,dx=\int_0^{\frac {\pi}2} \log(2\sin (\frac x2)\cos (\frac x2)))\,dx$$ $$=\frac {\pi}2\log(2)+2\int_0^{\frac{\pi}4} \log(\sin (y))+\log(\cos (y))\,dy$$ $$=\frac {\pi}2\log(2)+2\int_0^{\frac{\pi}4} \log(\sin (y))dy+2\int_0^{\frac{\pi}4}\log(\cos (y))\,dy$$ establecimiento $z=\frac {\pi}2-y$ a la derecha : $$=\frac {\pi}2\log(2)+2\int_0^{\frac{\pi}4} \log(\sin (y))dy-2\int_{\frac {\pi}2}^{\frac{\pi}4}\log(\sin (z))\,dz$$ $$\int_0^{\frac {\pi}2} \log(\sin (x))\,dx=\frac {\pi}2\log(2)+2\int_0^{\frac{\pi}2} \log(\sin (y))dy$$

de modo que $$\int_0^{\frac {\pi}2} \log(\sin (x))\,dx=-\frac {\pi}2\log(2)$$ y su integral inicial debe ser :

$$\boxed{\displaystyle\int_0^{\frac {\pi}2} \frac x{\tan x}\, dx=\frac {\pi}2\log 2}$$

Answer 3

Tomemos $t=\tan x$

$$I= \int_0^{\frac{\pi}{2}} \frac{x}{\tan x} dx=\int_{0}^{\infty}\frac{\arctan t}{t(1+t^2)}dt $$ Considerar

$$I(a)= \int_{0}^{\infty}\frac{\arctan (at)}{t(1+t^2)}dt $$

$$\frac{dI}{da}=\int_{0}^{\infty}\frac{1}{(1+a^2t^2)(1+t^2)}dt =\frac{\pi}{2(1+a)}$$ y

$$I(a)=\frac{\pi}{2}\ln (1+a)+const$$ $const=0$ because $I(0)=0$

Así

$$I(a)=\frac{\pi}{2}\ln (1+a)$$ and $I=I(1)=\frac{\pi}{2}\ln (2)$