Studying the improper integral $\int^{1}_{0}\frac{\ln(x)}{\sqrt{1-x}}dx$ I want to study this integral:
$$\int^{1}_{0}\frac{\ln(x)}{\sqrt{1-x}}\mathrm dx$$
It is continuous on $(0,1)$, so we need to study its limits when $x$ tends to $0$ and $1$.
Let $$f\colon x \mapsto \frac{\ln(x)}{\sqrt{1-x}}$$
When $x \rightarrow 0$ we have $f(x) \sim \ln(x)$ and when $x \rightarrow 1$ we have $f(x) \sim 0$. So what can I conclude here? I know that is has to have finite limits at its borders, which isn't the case when $x \rightarrow 0$. So can I state the integral doesn't converge?
 A: Hint. You have nearly finished. Just observe that, as $a \to 0^+$,
$$
\int_a^1 \ln x \:dx=a - a \ln a-1 \to -1
$$ giving the convergence of the given integral, since as $x \to 1^-$, 
$$
\frac{\ln(x)}{\sqrt{1-x}}=\frac{\ln(1-(1-x))}{\sqrt{1-x}} \sim -\sqrt{1-x} \to 0.
$$
A: Using integration by parts and $x=1-t^2$
$$
\begin{align}
\int_0^1\frac{\log(x)}{\sqrt{1-x}}\,\mathrm{d}x
&=2\int_0^1\log(x)\,\mathrm{d}\!\left(1-\sqrt{1-x}\right)\\
&=-2\int_0^1\frac{1-\sqrt{1-x}}x\,\mathrm{d}x\\
&=-2\int_0^1\frac1{1+\sqrt{1-x}}\,\mathrm{d}x\\
&=-4\int_0^1\frac{t}{1+t}\,\mathrm{d}t\\
&=-4\int_0^1\left(1-\frac1{1+t}\right)\,\mathrm{d}t\\[6pt]
&=-4(1-\log(2))
\end{align}
$$
A: In a right neighbourhood of the origin, $\frac{\log x}{\sqrt{1-x}}$ behaves like $\log x$, that is an integrable function, since:
$$ \int_{0}^{u}\log(x)\,dx = \left.x\left(\log x-1\right)\right|_{0^+}^{u}$$
and $\lim_{x\to 0^+}x\log(x)=0$. In a left neighbourhood of $x=1$, by setting $y=1-x$ we have that the integrand function behaves like $\frac{\log(1-y)}{\sqrt{y}} = -\sqrt{y}+O(y)$, that is continuous. So the integral exists and it is given by
$$ \frac{d}{d\alpha}\left.\int_{0}^{1} x^\alpha(1-x)^{-1/2}\,dx\right|_{\alpha=0^+}=\frac{d}{d\alpha}\left.\frac{\Gamma(\alpha+1)\Gamma(1/2)}{\Gamma(\alpha+3/2)}\right|_{\alpha=0^+}\stackrel{(*)}{=}2\left(\psi(1)-\psi(3/2)\right) $$
where $(*)$ follows from $\Gamma'(x)=\Gamma(x)\psi(x)$. Simplifying,
$$ \int_{0}^{1}\frac{\log x}{\sqrt{1-x}}\,dx = -\sum_{n\geq 0}\frac{4}{(2n+2)(2n+3)}=\color{blue}{4\,(\log 2-1)}.$$
