Euler integral - beta function

Question

$$ \int_0^\infty {\frac{x^{p-1} \ln x} {1+x}}\,dx $$ I have a solution but I really do not undestand why it is so. I know that one form of beta function is $$ B(x,y)=\int_0^\infty {\frac{t^{x-1}} {(1+t)^{x+y}}}\,dt $$ and if y=1-x it means $$ B(x,1-x)=\int_0^\infty {\frac{t^{x-1}} {1+t}}\,dt = {\frac{\pi} {sin\pi x}}\ $$ And our teacher has written down this $$ \int_0^\infty {\frac{x^{p-1} \ln x} {1+x}}\,dx = {\frac{d} {d p}}\ B(p,1-p)={-\frac{\pi^2cos(\pi p)} {sin^2(\pi p)}}$$ Thank you for any help!

Answer 1

Using derivative under the integral sign with $f$ the function defined as $$ f\left(p,x\right)=\frac{x^{p-1}}{1+x}=\frac{e^{\left(p-1\right)\ln\left(x\right)}}{1+x} $$ Then $$ \frac{\partial f }{\partial p}\left(p,x\right)=\frac{\ln\left(x\right)e^{\left(p-1\right)\ln\left(x\right)}}{1+x}=\frac{x^{p-1}\ln\left(x\right)}{1+x} $$ Then, justifying the derivative under the integral sign $$ \left(B\left(p,1-p\right)\right)'=\left(\int_{0}^{+\infty}f\left(p,x\right)\text{d}x\right)'=\int_{0}^{+\infty}\frac{\partial f }{\partial p}\left(p,x\right)\text{d}x=\int_{0}^{+\infty}\frac{x^{p-1}\ln\left(x\right)}{1+x}\text{d}x $$ And differenciating $\displaystyle x \mapsto \frac{\pi}{\sin\left(\pi x \right)}$ we have $$

$$ \left(B\left(p,1-p\right)\right)'=\int_{0}^{+\infty}\frac{x^{p-1}\ln\left(x\right)}{1+x}\text{d}x=-\pi^2\frac{\text{cotan}\left(\pi p\right)}{\sin\left(\pi p\right)} $$