Calculate $\int_0^{1}\frac{x^{-1 - x}\,\,\,\left(1 - x\right)^{x - 2}}{\mathrm{B}(1 - x\,, \,x)}\,\mathrm{d}x$ How does one calculate
$\displaystyle{\int_0^{1}\frac{x^{-1 - x}\,\,\,\left(1 - x\right)^{x - 2}}
{\mathrm{B}(1 - x\,,\,x)}\,\mathrm{d}x}$ ?.
The observation
$\displaystyle{\int_0^{1}\frac{x^{-1 - x}\,\,\,\left(1 - x\right)^{x - 2}}{\Gamma\left(1 - x\right)\Gamma\left(x\right)}\,\mathrm{d}x =
\int_0^{1}\frac{x^{-1 - x}\,\,\,\left(1 - x\right)^{x - 2}}
{\mathrm{B}(1 - x\,,\,x)}\,\mathrm{d}x}$ seems useless here. 
The answer, according to wolfram, is $2$. 

For the proof, we will be using the following fact $$(1-\alpha)^{\left(\frac{1}{\alpha}-1\right)} = \int_0^1 \frac{\sin\left(\pi x \right)}{\pi (1-\alpha x)}  \left[x^x\left(1-x\right)^{1-x}\right]^{-1}\ dx.$$
Taking your original sum, we have
\begin{align}
\int_0^1 \frac{x^{-1-x}(1-x)^{x-2}}{\Gamma(x)\Gamma(1-x)} dx &= \int_0^1 \frac{\sin(\pi x)}{\pi x(1-x)} \left[x^x\left(1-x\right)^{1-x}\right]^{-1} \ dx \\
&= \int_0^1 \frac{\sin(\pi x)}{\pi} \left[x^x\left(1-x\right)^{1-x}\right]^{-1} \left( \frac{1}{x}+\frac{1}{1-x}\right)\ dx \\
&=2\int_0^1 \frac{\sin(\pi x)}{\pi (1-x)} \left[x^x\left(1-x\right)^{1-x}\right]^{-1} \ dx \\
&= 2\lim_{\alpha \to 1^-} (1-\alpha)^{\left(\frac{1}{\alpha}-1\right)} =2\times 1 = 2
\end{align}
