How to prove using the Beta function of form
$$\mathrm{B}(a,b)=\int_0^\infty \frac{x^{a-1}}{(1+x)^{a+b}}\ dx\tag1$$
that
$$P=\int_0^\infty\frac{\ln^3x\ln(1+x)}{x(1+x)}\ dx=6\zeta(2)\zeta(3)+6\zeta(5)\ ?$$
This integral is already calculated here but I want it calculated without converting the boundaries $(0,\infty)$ to other boundaries and by using $(1)$ exclusively.
What I tried is
$$\frac{\partial }{\partial b}\mathrm{B}(a,b)=-\int_0^\infty \frac{\ln(1+x)x^{a-1}}{(1+x)^{a+b}}\ dx$$
set $b=0$
$$\frac{\partial }{\partial b}\mathrm{B}(a,b)|_{b\to 0}=-\int_0^\infty \frac{\ln(1+x)x^{a-1}}{(1+x)^{a}}\ dx$$
and if I differentiate again but with respect to $a$ three times, I will get $\ln^3(1+x)$ besides $\ln^3x$ and eventually I will have $\ln^4(1+x)$ in the numerator, so how to get the integrand of $P\ $?
Thank you
Best Answer
You could shift the second variable in $(1)$ to obtain $$ \operatorname{B}(a,b-a) = \int \limits_0^\infty \frac{x^{a-1}}{(1+x)^b} \, \mathrm{d} x \,.$$ Then $$ \frac{\partial^3}{\partial a^3} \frac{\partial}{\partial b} \operatorname{B}(a,b-a) = - \int \limits_0^\infty \frac{\log^3(x) \log(1+x) x^{a-1}}{(1+x)^b} \, \mathrm{d} x \,,$$ so (by analytic continuation) \begin{align} P &= - \frac{\partial^3}{\partial a^3} \frac{\partial}{\partial b} \operatorname{B}(a,b-a) \, \Bigg \rvert_{a=0, \, b=1} = - \frac{\partial^3}{\partial a^3} \operatorname{\Gamma}(a) \frac{\partial}{\partial b} \frac{\operatorname{\Gamma}(b-a)}{\operatorname{\Gamma}(b)} \, \Bigg \rvert_{a=0,\, b=1} \\ &= - \frac{\partial^3}{\partial a^3} \operatorname{\Gamma}(a) \operatorname{\Gamma}(1-a) [\operatorname{\psi}_0 (1-a) + \gamma] ~\Bigg \rvert_{a=0} = - \frac{\partial^3}{\partial a^3} \frac{\pi}{\sin(\pi a)} [\operatorname{\psi}_0 (1-a) + \gamma] ~\Bigg \rvert_{a=0} \\ &= - \frac{1}{4} [2 \pi^2 \operatorname{\psi}_2(1) + \operatorname{\psi}_4 (1)] = \pi^2 \operatorname{\zeta}(3) + 6 \operatorname{\zeta(5)} = 6[\operatorname{\zeta(2)} \operatorname{\zeta}(3) + \operatorname{\zeta(5)}] \, . \end{align}