Problem with evaluating $\int_0^{\frac{\pi}{2}} \ln(\sin(\theta))d\theta$ using Beta Function Recently I've been trying to tackle the integral $\int_0^{\frac{\pi}{2}} \ln(\sin(\theta))d\theta$ using the Beta function
$$\frac{B(\frac{x}{2},\frac{1}{2})}{2}=\int_0^{\frac{\pi}{2}} \sin^{x-1}(\theta)d\theta=\frac{\sqrt{\pi}}{2}\left(\Gamma\left(\frac{x+1}{2}\right)\right)^{-1}$$
Differentiating both sides
$$\int_0^{\frac{\pi}{2}} \ln(\sin(\theta))\sin^{x-1}(\theta)d\theta=-\frac{\sqrt{\pi}}{4}\frac{\psi(\frac{x+1}{2})}{\Gamma(\frac{x+1}{2})}$$
However, at $x=1$ $$\int_0^{\frac{\pi}{2}} \ln(\sin(\theta))d\theta\ne\frac{\gamma\sqrt{\pi}}{4}$$
Where did I go wrong?
 A: You forgot the $\Gamma(x/2)$ factor.
A: Cleaner approach: for any $\alpha\geq 0$,
$$ \int_{0}^{\pi/2}\left(\sin\theta\right)^{\alpha}\,d\theta = \int_{0}^{1}\frac{u^\alpha}{\sqrt{1-u^2}}\,du = \frac{1}{2}\int_{0}^{1}v^{\frac{\alpha-1}{2}}(1-v)^{-1/2}\,dv = \frac{\Gamma\left(\frac{\alpha+1}{2}\right)}{\Gamma\left(\frac{\alpha+2}{2}\right)}\cdot\frac{\sqrt{\pi}}{2}$$
Now we differentiate both sides with respect to $\alpha$. In order to differentiate the RHS, we multiply it by its logarithmic derivative. We get:
$$ \int_{0}^{\pi/2}\left(\sin\theta\right)^{\alpha}\log\sin\theta\,d\theta = \frac{\sqrt{\pi}}{4}\cdot \frac{\Gamma\left(\frac{\alpha+1}{2}\right)}{\Gamma\left(\frac{\alpha+2}{2}\right)}\cdot\left[\psi\left(\tfrac{\alpha+1}{2}\right)-\psi\left(\tfrac{\alpha+2}{2}\right)\right]. $$
Now we evaluate at $\alpha=0$, recalling that $\Gamma(1)=1,\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$ and 
$$ \sum_{n\geq 0}\frac{1}{(n+1)\left(n+\tfrac{1}{2}\right)}=\frac{\psi(1)-\psi(1/2)}{1-1/2}= 4\log 2.$$
The final outcome is:
$$ \int_{0}^{\pi/2}\log\sin\theta\,d\theta = -\frac{\pi}{2}\log 2.$$
A: Another way that only requires knowledge of the beta function (and expansion of the gamma function) is given as follows. Consider the integral
$$\begin{aligned}\int_{0}^{\pi/2}\sin^{\epsilon}x\,\mathrm{d}x &= \int_{0}^{\pi/2}e^{\epsilon\ln\sin x}\,\mathrm{d}x = \sum_{n=0}^{\infty}\frac{\epsilon^{n}}{n!}\int_{0}^{\infty}\ln^{n}\sin x\,\mathrm{d}x \\
&= \frac{\Gamma(1/2)\Gamma(1/2+\epsilon/2)}{2\,\Gamma(1+\epsilon/2)}. \end{aligned}$$ 
We have to match the appropriate coefficient of $\epsilon$ found by expanding the latter expression to the coefficient of the series in the first line. The integral we are interested in is the $n=1$ term. $\epsilon^{2}$ and higher order terms can then be neglected. Due to the trigonometric form of the beta function, integrals of this type often require the use of Legendre's duplication formula
$$ \frac{\Gamma(1+\epsilon)}{\Gamma(1/2+\epsilon)} = \frac{2^{2\epsilon}}{\sqrt{\pi}}\frac{\Gamma^{2}(1+\epsilon)}{\Gamma(1+2\epsilon)}, $$
derived also by using the beta function. The expansion of the gamma function around $1$ is given as
$$ \ln\Gamma(1+\epsilon) = -\gamma\epsilon + \sum_{k=2}^{\infty}\frac{(-1)^{k}\zeta(k)}{k}\epsilon^{k} $$
which is highly useful. Here $\gamma$ is the Euler-Mascheroni constant and $\zeta(k)$ is the Riemann zeta function. Note that the form of the expansion of the gamma function and the ratio of gamma functions $\Gamma(1+\epsilon)/\Gamma^{2}(1+\epsilon/2)$ means that the first order term of this ratio cannot contribute (it is just $1$), so up to first order,
$$\begin{aligned} \frac{\Gamma(1/2)\Gamma(1/2+\epsilon/2)}{2\,\Gamma(1+\epsilon/2)} &= \frac{\sqrt{\pi}}{2}\frac{\sqrt{\pi}}{2^{\epsilon}}\frac{\Gamma(1+\epsilon)}{\Gamma^{2}(1+\epsilon/2)} \\
&\approx \frac{\pi}{2}\frac{1}{e^{(\ln 2)\epsilon}} \approx \frac{\pi}{2}(1 - (\ln 2)\epsilon). \end{aligned}$$
In general, if we look for the coefficient of the $n$th term, we have to multiply by $n!$ to account for the factorial in the original expansion. First order term is trivial: $1! = 1$, so the first order coefficient is the answer
$$ \int_{0}^{\pi/2}\ln\sin x\,\mathrm{d}x = -\frac{\pi}{2}\ln 2. $$
