Using differentiation under integral sign, find the value of $\int_{0}^{1} \frac {x^{\alpha } + x^{-\alpha} -2}{(1-x) \ln(x)} dx$ Using differentiation under integral sign, find the value of $\int_{0}^{1} \dfrac {x^{\alpha } + x^{-\alpha} -2}{(1-x) \ln(x)} dx$
EDIT:My Attempt after P. Lawrence's comment:
The given integral is : $\int_{0}^{1} \dfrac {x^{\alpha } + x^{-\alpha} -2}{(1-x) \ln(x)} dx$.
Here , $\alpha $ is the parameter so we let:
$$F(\alpha )=\int_{0}^{1} \dfrac {x^{\alpha } + x^{-\alpha} -2}{(1-x) \ln(x)} dx$$
Differentiating both sides wrt $\alpha $
$$\dfrac {dF(\alpha )}{d\alpha }=\dfrac {d}{d\alpha} \int_{0}^{1} \dfrac {x^{\alpha } + x^{-\alpha} -2}{(1-x) \ln(x)} dx$$
By Leibnitz Theorem:
$$\dfrac {dF(\alpha) }{d\alpha} = \int_{0}^{1} \dfrac {x^{\alpha}\ln(x)-x^{-\alpha}\ln(x)}{(1-x)\ln(x)} dx$$
$$\dfrac {dF(\alpha )}{d\alpha }=\int_{0}^{1} \dfrac {x^{\alpha} - x^{-\alpha}}{1-x} dx$$ 
 A: After OP's work, Here $H_x=\int_{0}^{1} \frac{t^x-1}{x-1} dt$ are the well known Harmonic numcers. Like $H_n=1+1/2+1/3+1/4+...+1/n$
$$\frac{dF(a)}{da}=-\int_{0}^{1} \frac{x^a-x^{-a}}{x-1} dx= -\int_{0}^{1} \frac{(x^a-1)-(x^{-a}-1)}{x-1} dx=- H_a+H_{-a}$$
Then 
$$F(a)=\int (-H_a+H_{-a}) da =-\ln\Gamma(1+a)-\ln\Gamma(1-a)+C$$
We have used $\int H_x dx=\gamma+\ln \Gamma(1+x).$
As $F(0)=0 \implies C=0.$
Hence the required summation is finite if $0<a<1$ and it is given by
$$S=-[\ln \Gamma(1+a)+\ln\Gamma(1-a) =-\ln[ \Gamma(1-a) \Gamma(1+a)]=\ln \frac{\sin \pi a}{\pi a}$$
Here we have use $\Gamma(1-z) \Gamma(1+z)=\frac{\pi z}{\sin \pi z}$
For Harmonic numbers one may see:
https://en.wikipedia.org/wiki/Harmonic_number
A: Calling the integral $f(\alpha)$, differentiating, and using $(1-x)^{-1} = \sum x^k$ for $|x| < 1$, we have
\begin{aligned} \displaystyle f'(\alpha)  &= \int_0^1 \frac{x^\alpha-x^{-\alpha}}{1-x}\,\mathrm{d}x  = \int_0^1 (x^\alpha-x^{-\alpha}) \sum_{k \ge 0}x^k \,\mathrm{d}x \\& \overset{\text{D.C.T}}= \sum_{k \ge 0} \int_0^1 (x^{k+\alpha}-x^{k-\alpha}) \,\,\mathrm{d}x  = \sum_{ k \ge 0} \left[\frac{1}{k+\alpha+1}- \frac{1}{k-\alpha+1}\right] \\& = \sum_{k \ge 0} \frac{2\alpha}{\alpha^2-(k+1)^2}  = \sum_{k \ge 1} \frac{2\alpha}{\alpha^2-k^2}=\pi  \cot(\pi \alpha)-\frac{1}{\alpha }. \end{aligned}
Where the last step is the cotangent series $\displaystyle \pi \cot(\pi z) = \frac{1}{z}+2z \sum_{n \ge 1}\frac{1}{z^2-n^2}.$
Thus $\displaystyle f'(\alpha) = \pi \cot(\pi \alpha)-\frac{1}{\alpha }$, hence $\displaystyle f(\alpha) = \log \left(\frac{\sin  \pi \alpha }{\alpha }\right)+k$ for some $k \in \mathbb{R}$. 
We know from the integral that  $f(0) = 0$ and from our antiderivative that $f(\alpha) \to k+\log(\pi)$ as $\alpha \to 0$. Therefore $k = - \log(\pi)$ hence $$\displaystyle f(\alpha) = \log \left(\frac{\sin  \pi \alpha }{\pi \alpha }\right).$$
