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

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 $1/2-a>-1.$
