Question-162864

Question Number 162864 by mathacek last updated on 01/Jan/22

Answered by Ar Brandon last updated on 01/Jan/22

I = \int_{0}^{π} x \ln (\sin x) d x = \int_{0}^{π} (π - x) \ln (\sin x) d x

= π \int_{0}^{π} \ln (\sin x) d x - \int_{0}^{π} x \ln (\sin x) d x

I = \frac{π}{2} \int_{0}^{π} \ln (\sin x) d x, x = 2 u \Rightarrow d x = 2 d u

= π \int_{0}^{\frac{π}{2}} \ln (\sin (2 u)) d x = π \int_{0}^{\frac{π}{2}} \ln (2 \sin u \cos u) d u

= π \int_{0}^{\frac{π}{2}} \ln (2) d u + π \int_{0}^{\frac{π}{2}} \ln (\sin u) d u + π \int_{0}^{\frac{π}{2}} \ln (\cos u) d u

= \frac{π^{2} \ln 2}{2} + 2 π \int_{0}^{\frac{π}{2}} \ln (\sin u) d u, {\int_{0}^{\frac{π}{2}} \ln (\sin u) d u = \int_{0}^{\frac{π}{2}} \ln (\cos u) d u}

= \frac{π^{2} \ln 2}{2} - π^{2} \ln 2 = - \frac{1}{2} π^{2} \ln 2

Commented by Ar Brandon last updated on 01/Jan/22

I = \int_{0}^{\frac{π}{2}} \ln (\sin x) d x = \int_{0}^{\frac{π}{2}} \ln (\cos x) d x

2 I = \int_{0}^{\frac{π}{2}} \ln (\sin x \cos x) d x = \int_{0}^{\frac{π}{2}} \ln (\frac{1}{2} \sin (2 x)) d x

= \int_{0}^{\frac{π}{2}} \ln (\frac{1}{2}) d x + \int_{0}^{\frac{π}{2}} \ln (\sin 2 x) d x

= - \frac{π \ln 2}{2} + \frac{1}{2} \int_{0}^{π} \ln (\sin u) d u

Answered by Lordose last updated on 01/Jan/22

I = \int_{0}^{π} xln (\sin (x)) dx \overset{{King}^{'} s Rule}{=} \int_{0}^{π} (π - x) \ln (\sin (x)) dx

2 I = π \int_{0}^{π} \ln (\sin (x)) dx \overset{u = 2 x}{=} \frac{π}{2} \int_{0}^{2 π} \ln (\sin (\frac{u}{2})) dx

I = \frac{π}{4} (- {Cl}_{2} (2 π) - 2 π \ln (2))

I = - \frac{π^{2} \ln (2)}{2}

Commented by mathacek last updated on 02/Jan/22

{W h a t}^{'} s f o r {Cl}_{2} ?

Commented by Ar Brandon last updated on 02/Jan/22

Clausen function

{C l}_{2} (z) = \sum_{n ⩾ 1} \frac{\sin (n z)}{n^{2}}