Work out $\int_{0}^{\pi/2}\cos^{2n}(x)\ln\sin(x) dx$

Question

$$F(n)=\large \int_{0}^{\pi/2}\cos^{2n}(x)\ln\sin(x)\mathrm dx$$

$$\cos^{2n}(x)=\frac{1}{2^{2n}}{2n \choose n}+\frac{1}{2^{2n-1}}\sum_{k=0}^{n-1}{2n \choose k}\cos\left[2\left(n-k\right)x\right]$$

$$I=\frac{1}{2^{2n}}{2n \choose n}\int_{0}^{\pi/2}\ln\sin(x)\mathrm dx+\frac{1}{2^{2n-1}}\sum_{k=0}^{n-1}{2n \choose k}\int_{0}^{\pi/2}\cos\left[2\left(n-k\right)x\right]\ln\sin(x)\mathrm dx$$

$$I=-\frac{\pi \ln(2)}{2^{2n+1}}{2n \choose n}+\frac{1}{2^{2n-1}}\sum_{k=0}^{n-1}{2n \choose k}\int_{0}^{\pi/2}\cos\left[2\left(n-k\right)x\right]\ln\sin(x)\mathrm dx$$

I can't continue.

I would like to evaluate $F(n)$

Answer 1

You can let $\cos^2 (x) = t$ to obtain \begin{align} F(n) &= \frac{1}{4} \int \limits_0^1 t^{n-\frac{1}{2}} (1-t)^{-\frac{1}{2}} \ln(1-t) \, \mathrm{d} t \\ &= \frac{1}{4}\frac{\mathrm{d}}{\mathrm{d}s} \int \limits_0^1 t^{n-\frac{1}{2}} (1-t)^{s - 1} \, \mathrm{d} t ~\Bigg \lvert_{s=\frac{1}{2}} \\ &= \frac{1}{4} \frac{\mathrm{d}}{\mathrm{d}s} \operatorname{B} \left(n+\frac{1}{2},s\right)\\ &= -\frac{\Gamma\left(n+\frac{1}{2}\right) \Gamma\left(\frac{1}{2}\right)}{4\Gamma(n+1)} \left[\psi (n+1) - \psi \left(\frac{1}{2}\right)\right] \, . \end{align} For $n \in \mathbb{N}_0$ the special values $\Gamma\left(n+\frac{1}{2}\right) = \frac{(2n)! \sqrt{\pi}}{4^n n!}$ , $\psi \left(\frac{1}{2}\right) = - \gamma - 2 \ln(2)$ and $\psi(n+1) = H_n - \gamma$ yield $$F(n) = - \frac{\pi}{4} \frac{(2n-1)!!}{(2n)!!}[H_n + 2 \ln(2)] \, . $$

Answer 2

Note that for $x\in (0,\pi/2]$, $$\ln(\sin(x))=\ln\left(\frac{e^{ix}-e^{-ix}}{2i}\right) =-\ln(2)-\sum_{j=1}^\infty\frac{\cos(2jx)}{j}.$$ Hence, for $0\leq k<n$ and $j\geq 1$, $$\int_{0}^{\pi/2}\cos(2\left(n-k\right)x)\cos(2jx) dx=\begin{cases} \pi/4 &\text{if $n-k=j$}\\ 0&\text{otherwise} \end{cases}.$$ Therefore, following your approach, we obtain that $$\begin{align*} F(n)&=-\frac{\pi}{2^{2n+1}}\left(\ln(2)\binom{2n}{n}+\sum_{k=0}^{n-1}\binom{2n}{k}\frac{1}{n-k}\right)\\ &=-\frac{\pi}{2^{2n+2}}\binom{2n}{n}\left(2\ln(2)+H_n\right)\end{align*}.$$

Answer 3

Hint: With $\beta$-function $$\beta(x,y) = 2\int_{0}^{\pi/2}\sin^{2x-1}\theta \cos^{2y-1}\theta d\theta = \dfrac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$ then $$\dfrac{d}{dx}\beta(x,n+\frac12) =\dfrac{d}{dx} 2\int_{0}^{\pi/2}\sin^{2x-1}\theta \cos^{2n}\theta d\theta=\int_{0}^{\pi/2}\sin^{2x-1}\theta \cos^{2n}\theta \ \ln\sin x d\theta$$ and $$F(n)=\left(\dfrac{d}{dx}\beta(x,n+\frac12)\right)_{x=\frac12} =\left(\dfrac{d}{dx}\dfrac{\Gamma(x)\Gamma(n+\frac12)}{\Gamma(x+n+\frac12)}\right)_{x=\frac12} $$