Prove-that-0-tln-sint-dt-2-2-ln-2-

Question Number 146775 by Willson last updated on 15/Jul/21

Prove that

{\int_{0}}^{π} t l n (s i n t) d t = - \frac{π^{2}}{2} l n (2)

Answered by ArielVyny last updated on 15/Jul/21

\int_{0}^{π} t (- l n 2 - \sum_{n ⩾ 1} \frac{c o s (2 n t)}{n}) d t

- l n (2) \int_{0}^{π} t d t - \int_{0}^{π} (t \sum_{n ⩾ 1} \frac{c o s (2 n t)}{n}) d t

- \frac{π^{2}}{2} l n 2 a c c o r d i n g t h a t \int_{0}^{π} t \sum_{n ⩾ 1} \frac{c o s (2 n t)}{n} d t = 0

\int_{0}^{π} t l n (s i n t) d t = - \frac{π^{2} l n 2}{2}

Answered by Olaf_Thorendsen last updated on 15/Jul/21

Ω = \int_{0}^{π} t \ln (\sin t) d t (1)

Let u = π - t

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

\frac{(1) + (2)}{2} : Ω = \frac{π}{2} \int_{0}^{π} \ln (\sin t) d t (3)

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

Let u = t - \frac{π}{2}

Ω = \frac{π}{2} \int_{0}^{\frac{π}{2}} \ln (\sin t) d t + \frac{π}{2} \int_{0}^{\frac{π}{2}} \ln (\cos u) d u (4)

Let t = 2 u

(3) : Ω = π \int_{0}^{\frac{π}{2}} \ln (\sin (2 u)) d u

Ω = π \int_{0}^{\frac{π}{2}} \ln 2 + \ln (\sin u) + \ln (\cos u) d u (5)

(5) - 2 \times (4) w :

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

\Rightarrow Ω = - \frac{π^{2}}{2} \ln (2)

Answered by mathmax by abdo last updated on 16/Jul/21

Υ = \int_{0}^{π} t \log (sint) dt \Rightarrow Υ = \int_{0}^{\frac{π}{2}} tlog (sint) dt + \int_{\frac{π}{2}}^{π} tlog (sint) dt (\to t = \frac{π}{2} + z)

= \int_{0}^{\frac{π}{2}} tlog (sint) dt [+ \int_{0}^{\frac{π}{2}} (\frac{π}{2} + z) \log (cost) dt

= \frac{π}{2} \int_{0}^{\frac{π}{2}} \log (cost) dt + \int_{0}^{\frac{π}{2}} t (\log (sint) + \log (cost)) dt

= - \frac{π^{2}}{4} \log 2 + \int_{0}^{\frac{π}{2}} tlog (\frac{\sin (2 t)}{2}) dt

= - \frac{π^{2}}{4} \log 2 + \int_{0}^{\frac{π}{2}} tlog (\sin (2 t) dt (\to 2 t = u) - \log 2 \int_{0}^{\frac{π}{2}} tdt

= - \frac{π^{2}}{4} \log 2 + \int_{0}^{π} \frac{u}{2} \log (sinu) \frac{du}{2} - \log 2 \times \frac{π^{2}}{8}

= - \frac{3 π^{2}}{8} \log 2 + \frac{Υ}{4} \Rightarrow (1 - \frac{1}{4}) Υ = - \frac{3 π^{2}}{8} \log 2 \Rightarrow

\frac{3}{4} Υ = - \frac{3 π^{2}}{8} \log 2 \Rightarrow Υ = - \frac{π^{2}}{2} \log 2