Csdrhrt

Recently I've been trying to tackle the integral $int_0^{frac{pi}{2}} ln(sin(theta))dtheta$ using the Beta function
$$frac{B(frac{x}{2},frac{1}{2})}{2}=int_0^{frac{pi}{2}} sin^{x-1}(theta)dtheta=frac{sqrt{pi}}{2}left(Gammaleft(frac{x+1}{2}right)right)^{-1}$$
Differentiating both sides
$$int_0^{frac{pi}{2}} ln(sin(theta))sin^{x-1}(theta)dtheta=-frac{sqrt{pi}}{4}frac{psi(frac{x+1}{2})}{Gamma(frac{x+1}{2})}$$
However, at $x=1$ $$int_0^{frac{pi}{2}} ln(sin(theta))dthetanefrac{gammasqrt{pi}}{4}$$

Where did I go wrong?

You forgot the $Gamma(x/2)$ factor.

Cleaner approach: for any $alphageq 0$,

$$ int_{0}^{pi/2}left(sinthetaright)^{alpha},dtheta = int_{0}^{1}frac{u^alpha}{sqrt{1-u^2}},du = frac{1}{2}int_{0}^{1}v^{frac{alpha-1}{2}}(1-v)^{-1/2},dv = frac{Gammaleft(frac{alpha+1}{2}right)}{Gammaleft(frac{alpha+2}{2}right)}cdotfrac{sqrt{pi}}{2}$$
Now we differentiate both sides with respect to $alpha$. In order to differentiate the RHS, we multiply it by its logarithmic derivative. We get:

$$ int_{0}^{pi/2}left(sinthetaright)^{alpha}logsintheta,dtheta = frac{sqrt{pi}}{4}cdot frac{Gammaleft(frac{alpha+1}{2}right)}{Gammaleft(frac{alpha+2}{2}right)}cdotleft[psileft(tfrac{alpha+1}{2}right)-psileft(tfrac{alpha+2}{2}right)right]. $$
Now we evaluate at $alpha=0$, recalling that $Gamma(1)=1,Gammaleft(frac{1}{2}right)=sqrt{pi}$ and
$$ sum_{ngeq 0}frac{1}{(n+1)left(n+tfrac{1}{2}right)}=frac{psi(1)-psi(1/2)}{1-1/2}= 4log 2.$$
The final outcome is:
$$ int_{0}^{pi/2}logsintheta,dtheta = -frac{pi}{2}log 2.$$

Another way that only requires knowledge of the beta function (and expansion of the gamma function) is given as follows. Consider the integral

$$begin{aligned}int_{0}^{pi/2}sin^{epsilon}x,mathrm{d}x &= int_{0}^{pi/2}e^{epsilonlnsin x},mathrm{d}x = sum_{n=0}^{infty}frac{epsilon^{n}}{n!}int_{0}^{infty}ln^{n}sin x,mathrm{d}x \
&= frac{Gamma(1/2)Gamma(1/2+epsilon/2)}{2,Gamma(1+epsilon/2)}. end{aligned}$$

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

$$ frac{Gamma(1+epsilon)}{Gamma(1/2+epsilon)} = frac{2^{2epsilon}}{sqrt{pi}}frac{Gamma^{2}(1+epsilon)}{Gamma(1+2epsilon)}, $$

derived also by using the beta function. The expansion of the gamma function around $1$ is given as

$$ lnGamma(1+epsilon) = -gammaepsilon + sum_{k=2}^{infty}frac{(-1)^{k}zeta(k)}{k}epsilon^{k} $$

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

$$ int_{0}^{pi/2}lnsin x,mathrm{d}x = -frac{pi}{2}ln 2. $$