Calculate the integral
$$\int\limits_{0}^{1}\frac{x^{1-p}(1-x)^p}{(1+x)^2}dx, \; -1<p<2$$
My attempt: I tried to use the Laplace transform
$$\mathcal{L}\left \{ \frac{x^{1-p}(1-x)^p}{(1+x)^2} \right \}(s)=\int_0^1 \frac{x^{1-p}(1-x)^p}{(1+x)^2}e^{-sx}dx=\int_0^\infty e^{-sx}\frac{x^{1-p}(1-x)^p}{(1+x)^2}dx=\int_0^\infty e^{-sx}\int_0^\infty e^{-tu}\frac{(tu)^{1-p}[(1-t)+(1-u)]^p}{(1+tu)^2}dudt$$
I don't know what to do next. Is it possible to reduce it to a beta function? This integral is very similar to this function. But I haven't been able to find any substitute
Best Answer
Substitute $x=\frac1{1+y}$ \begin{align} I=&\int_{0}^{1}\frac{x^{1-p}(1-x)^p}{(1+x)^2}dx =\int_0^\infty\frac{y^p}{(1+y)(2+y)^2}dy\\ =& \int_0^\infty \frac{y^p}{1+y}dy-\int_0^\infty\frac{y^p}{2+y}dy- \int_0^\infty\frac{y^p}{(2+y)^2}dy \end{align} Utilize the known integral $J(a)=\int_0^\infty \frac {y^p}{a+y}dy= -\pi a^p \csc\pi p$ to evaluate \begin{align} & \int_0^\infty \frac{y^p}{1+y}dy=J(1)= -\pi \csc\pi p\\ & \int_0^\infty \frac{y^p}{2+y}dy=J(2)= -\pi \ 2^p\csc\pi p\\ & \int_0^\infty \frac{y^p}{(2+y)^2}dy=-\frac{d J(a)}{da}\bigg|_{a=2} = \pi \ 2^{p-1}p\csc\pi p\\ \end{align} Plug above results into $I$ to obtain $$I=\pi\ [2^{p-1}(2-p)-1]\csc\pi p $$