Convergence $\int_0^1{\log{(\arctan{}x)}\dfrac{\frac{\pi}{2}-\arcsin{x}}{(e^{1-x}-1)^a}}$ I have difficulties with solving the convergence of Newton inetgral:
$$\int_0^1\log{(\arctan{}x)}\dfrac{\frac{\pi}{2}-\arcsin{x}}{(e^{1-x}-1)^a}, a\in\mathbb{R}$$
Edit Show for which $a$ does it converges.
I don't know much where/how to start so I would appreciate any help.
My suggestions:
Divide it on two intervals, let's say $$\int_0^{1/2}\log{(\arctan{}x)}\dfrac{\frac{\pi}{2}-\arcsin{x}}{(e^{1-x}-1)^a}
+\int_{1/2}^1\log{(\arctan{}x)}\dfrac{\frac{\pi}{2}-\arcsin{x}}{(e^{1-x}-1)^a}$$
In the first interval, using limit comparison test with function $g(x)=\frac{1}{\sqrt{x}}$, we get 
$$lim_{x\rightarrow0}\frac{f(x)}{g(x)}=\cdots=0$$ And since $\int_0^{1/2}\frac{1}{\sqrt{x}}$ converges, we have, that$\int_0^{1/2}\log{(\arctan{}x)}\dfrac{\frac{\pi}{2}-\arcsin{x}}{(e^{1-x}-1)^a}$ converges for all $a$.
But the second part seems to be trickier.
Edit 2:
Is it possible that we for the second part we will use limit comparison test with function $g(x)=(x-1)^{-a+1}$? Then we will get that it converges for all $a<1$...but it might be wrong
 A: Let $I(a)$ be the integral
$$I(a)=\int_0^1\,\log(\arctan x)\frac{\pi/2-\arcsin x}{(e^{1-x}-1)^a} \tag1$$
where $a$ is a real-valued parameter.
Note that the integrand of $(1)$ has singularities at both $x=0$ and $x=1$.  We shall investigate these singularities to determine when the integral in $(1)$ exists as an improper Riemann integral.  
First, the singularity at $x=0$ arises in the term $\log(\arctan x)$.  Noting that $\arctan x =x+O(x^3)$, we have 
$$\begin{align}
\log \arctan x &=\log (x+O(x^3))\\\\
&=\log x+\log(1+O(x^2))\\\\
&=\log x+O(x^2)
\end{align}$$
Inasmuch as $\int_0^1 \log x$ exists as an improper Riemann integral, we conclude that the singularity at $x=0$ does not render the integral divergent.  However, we still need to check the singularity at $x=1$.
The singularity at $x=1$ arises for $a>0$ due to the term $\frac{1}{(e^{1-x}-1)^a}$.  This singularity is of order $(1-x)^{-a}$ for $a>0$.  More precisely, 
$$\frac{1}{(e^{1-x}-1)^a}=(1-x)^{-a}(1+O(1-x))^{-a}$$
We also have (see NOTE)
$$\pi/2-\arcsin x=\sqrt{2}\sqrt{1-x}+O(1-x)^{3/2} \tag 2$$
so that 
$$\frac{\pi/2 -\arcsin x}{(e^{1-x}-1)^a}=\left(\sqrt{2}\sqrt{1-x}+O(1-x)^{3/2}\right)\left((1-x)^{-a}[1+O(x-1)]^{-a}\right)\tag 3$$
The leading term of $(3)$ is of order $(1-x)^{1/2-a}$.  Thus, the singularity at $x=1$ will lead to a convergent improper Riemann integral for values $a<3/2$.  Otherwise the integral diverges.

NOTE: On the Expansion of $\arcsin(x)$ Around $x=1^{-}$.
Let $f(x)=\arcsin(x)$.  Then, $f'(x)=\frac{1}{\sqrt{1-x^2}}$.  Therefore, $f'(x)\sim \frac{\sqrt 2}{2}\frac{1}{\sqrt{1-x}}$ near $x=1^{-}$.  
This motivates enforcing the substitution $x=1-y^2$ so that $y=\sqrt{1-x}$.  Note that as as $x\to 1^{-}$, $y\to 0^{+}$.  
We proceed now to expand $f(x)=f(1-y^2)=\arcsin(1-y^2)$ in a series around $y=0$.  Taking the first few derivatives of $f(1-y^2)$ with respect to $y$ gives for $y\ge 0$
$$\begin{align}
\frac{df}{dy}&=-2\left(2-y^2\right)^{-1/2}\\\\
\frac{d^2f}{dy^2}&=-2y\left(2-y^2\right)^{-3/2}\\\\
\frac{d^3f}{dy^3}&=-4(y^2+1)\left(2-y^2\right)^{-5/2}\\\\
\end{align}$$
Evaluating these derivatives at $y=0^+$ provides the expansion coefficients for the expansion of $\arcsin(1-y^2)$
$$\arcsin(1-y^2)=\frac{\pi}{2}-\sqrt 2 y-\frac{\sqrt 2}{12}y^3+O(y^5)$$
and therefore we have
$$\arcsin(x)=\frac{\pi}{2}-\sqrt 2 \sqrt{1-x}\left(1+\frac12 (1-x)+O(1-x)^2\right)$$
A: For the second interval, note first the $\log $ term has a finite nonzero limit at $1,$ so for matters of convergence we can ignore it. Second, recall $(e^h-1)/h\to 1$ as $h\to 0.$ This is just the definition of the derivative of $e^x$ at $0.$ It follows that in absolute value the denominator is on the order of $(1-x)^a.$ Upstairs we have
$$\frac{\pi}{2}- \arcsin x = \int_x^{1} \frac{dt}{\sqrt {1-t^2}}.$$
Now $\sqrt {1-t^2}$ is on the order of $\sqrt {1-t}$ if $t$ is near $1.$ You can integrate $1/\sqrt {1-t}$ easily to see the last integral is on the order of $(1-x)^{1/2}.$ It looks to me like your second integral converges iff $1/2-a>-1.$
