Mistake with Integration with Beta, Gamma, Digamma Fuctions Problem: Evaluate:

$$I=\int_0^{\pi/2} \ln(\sin(x))\tan(x)dx$$

I tried to attempt it by using the Beta, Gamma and Digamma Functions. My approach was as follows:
$$$$
Consider $$I(a,b)=\int_0^{\pi/2} \sin^a(x)\sin^b(x)\cos^{-b}(x)dx$$
$$$$
$$=\dfrac{1}{2}\beta\bigg (\dfrac{a+b+1}{2},\dfrac{1-b}{2}\bigg )= \dfrac{1}{2} \dfrac{\Gamma(\frac{a+b+1}{2})\Gamma(\frac{1-b}{2})}{\Gamma(\frac{a+2}{2})}$$ Now, 
$$\dfrac{1}{2}\dfrac{\partial}{\partial a} \beta\bigg (\dfrac{a+b+1}{2},\dfrac{1-b}{2}\bigg )\bigg |_{a=0,b=1} = \int_0^{\pi/2}\ln(\sin(x))\sin^a(x)\tan^b(x)dx \bigg |_{a=0,b=1} = I$$ 
Now, 
$$\dfrac{\partial}{\partial a} \beta\bigg (\dfrac{a+b+1}{2},\dfrac{1-b}{2}\bigg )$$
$$$$
$$=\dfrac{1}{2} \dfrac{\Gamma(\frac{1-b}{2})}{(\Gamma(\frac{a+2}{2}))^2}\bigg (\Gamma '\bigg (\frac{a+b+1}{2}\bigg )\Gamma \bigg ( \frac{a+2}{2}\bigg ) - \Gamma '\bigg ( \frac{a+2}{2}\bigg )\Gamma \bigg (\frac{a+b+1}{2}\bigg )\bigg ) $$
$$$$
$$= \dfrac{1}{2}\dfrac{\Gamma(\frac{1-b}{2})}{\Gamma(\frac{a+2}{2})}\bigg (\psi\bigg ( \frac{a+b+1}{2}\bigg )\Gamma \bigg (\frac{a+b+1}{2}\bigg ) - \psi\bigg ( \frac{a+2}{2}\bigg )\Gamma \bigg (\frac{a+b+1}{2}\bigg )\bigg )$$
$$$$
$$=\dfrac{1}{2} \dfrac{\Gamma(\frac{a+b+1}{2})\Gamma(\frac{1-b}{2})}{\Gamma(\frac{a+2}{2})}\bigg (\psi\bigg ( \frac{a+b+1}{2}\bigg )-\psi\bigg ( \frac{a+2}{2}\bigg )\bigg )$$
$$$$ 
$$\Rightarrow\dfrac{\partial}{\partial a} \beta\bigg (\dfrac{a+b+1}{2},\dfrac{1-b}{2}\bigg )=\dfrac{1}{2} \dfrac{\Gamma(\frac{a+b+1}{2})\Gamma(\frac{1-b}{2})}{\Gamma(\frac{a+2}{2})}\bigg (\psi\bigg ( \frac{a+b+1}{2}\bigg )-\psi\bigg ( \frac{a+2}{2}\bigg )\bigg )$$
$$$$
$$\Longrightarrow \dfrac{1}{2} \dfrac{\partial}{\partial a} \beta \bigg ( \dfrac{a+b+1}{2} ,\dfrac{1-b}{2} \bigg ) \bigg |_{a=0,b=1} = I $$
$$$$
$$=\dfrac{1}{2}\times \dfrac{1}{2} \dfrac{\Gamma(\frac{a+b+1}{2})\Gamma(\frac{1-b}{2})}{\Gamma(\frac{a+2}{2})}\bigg (\psi\bigg ( \frac{a+b+1}{2}\bigg )-\psi\bigg ( \frac{a+2}{2}\bigg )\bigg ) \bigg |_{a=0,b=1}$$
$$$$
$$ =\dfrac{1}{4}\dfrac{\Gamma(0)\Gamma(1)}{\Gamma(1)}\bigg (\psi\bigg ( 1 \bigg )-\psi\bigg (1\bigg )\bigg )$$
$$$$
Could somebody please be so kind as to tell me where I have gone wrong? I would be truly grateful for your assistance. Thanks very, very much in advance!
 A: OK, I realize this does not really answer your question (but look at the update below), in the sense it does not point at your error. Anyways, I think that you should be flexible with your methods doing integrals, so here is how it can be done with the change of variables I suggested: 
You end up with the integral
$$
I=\int_0^1 \log u\frac{u}{1-u^2}\,du.
$$
Now writing
$$
\frac{1}{1-u^2}=1+u^2+u^4+\cdots,
$$
and using that (integrating by parts)
$$
\begin{aligned}
\int_0^1 u^{2k+1}\log u\,du&=\bigl[\frac{u^{2k+2}}{2k+2}\log u\bigr]_0^1-\frac{1}{2k+2}\int_0^1 u^{2k+2}\frac{1}{u}\,du\\
&=-\frac{1}{4}\frac{1}{(k+1)^2}.
\end{aligned}
$$
Thus,
$$
I=-\frac{1}{4}\sum_{k=0}^{+\infty}\frac{1}{(k+1)^2}=-\frac{\pi^2}{24}.
$$
The last equality by the Basel problem.
Updated version
The problem in your calculation, by inserting $a=0$ and $b=1$, you get a $\Gamma(0)$ (which is undefined, or as its best $\pm\infty$ depending on if you let $b\to 1$ from left or right) times something that is zero. This is indeterminate, and you must look at limits. One way is as follows:
If you first let $a=0$, and write $b=1+\epsilon$, then you get
$$
\frac{1}{4}\Gamma(-\epsilon/2)\Gamma(1+\epsilon/2)\bigl(\Psi(1+\epsilon/2)-\Psi(1)\bigr)
$$
We multiply and divide by $\epsilon/2$ (lethal weapon number 2), to write this as
$$
\frac{1}{4}\frac{\epsilon}{2}\Gamma(-\epsilon/2)\Gamma(1+\epsilon/2)\frac{\Psi(1+\epsilon/2)-\Psi(1)}{\epsilon/2}
$$
By letting $\epsilon\to 0$, you will find that the limit is
$$
\frac{1}{4}\times (-1)\times 1 \times \frac{\pi^2}{6}=-\frac{\pi^2}{24}.
$$
Here, we have used the facts that $\lim_{x\to 0} x\Gamma(x)=1$ and $\Psi^{(1)}(1)=\pi^2/6$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
I & = \int_{0}^{\pi/2}\ln\pars{\sin\pars{x}}\tan\pars{x}\,\dd x
\,\,\,\stackrel{x\ =\ \arcsin\pars{t}}{=}\,\,\,
\int_{0}^{1}\ln\pars{t}{t \over \root{1 - t^{2}}}
\,{\dd t \over \root{1 - t^{2}}}
\\[5mm] & =
\int_{0}^{1}\ln\pars{t}\,{t \over 1 - t^{2}}\,\dd t =
\int_{0}^{1}\ln\pars{t}\pars{{1/2 \over 1 - t} - {1/2 \over 1 + t}}\,\dd t
\\[5mm] & =
-\,{1 \over 2}\int_{0}^{1}\bracks{-\,{{\ln\pars{1 - t} \over t}}}\,\dd t +
{1 \over 2}\int_{0}^{-1}{\ln\pars{-t} \over 1 - t}\,\dd t
\\[5mm] & =
-\,{1 \over 2}\int_{0}^{1}\bracks{-\,{{\ln\pars{1 - t} \over t}}}\,\dd t -
{1 \over 2}\int_{0}^{-1}\bracks{-\,{\ln\pars{1 - t} \over t}}\,\dd t
\\[5mm] & =
-\,{1 \over 2}\int_{0}^{1}\mrm{Li}_{2}'\pars{t}\,\dd t -
{1 \over 2}\int_{0}^{-1}\mrm{Li}_{2}'\pars{t}\,\dd t =
-\,{1 \over 2}\bracks{\mrm{Li}_{2}\pars{1} + \mrm{Li}_{2}\pars{-1}}
\\[5mm] & =
-\,{1 \over 2}\sum_{n = 1}^{\infty}
\bracks{{1 \over n^{2}} + {\pars{-1}^{n} \over n^{2}}} =
-\,{1 \over 2}\sum_{n = 1}^{\infty}
\bracks{{1 \over \pars{2n}^{2}} + {1 \over \pars{2n}^{2}}} =
-\,{1 \over 4}\ \underbrace{\sum_{n = 1}^{\infty}{1 \over n^{2}}}
_{\ds{\pi^{2} \over 6}} = \bbx{-\,{\pi^{2} \over 24}}
\end{align}
