I am going to evaluate the integral $$\displaystyle I=\int_{0}^{\infty} \frac{x^{a} \ln x}{1+x^{b}} d x, \tag*{} $$ where $b>a+1>0$, by its partner integral $$\displaystyle J(a)=\int_{0}^{\infty} \frac{x^{a}}{1+x^{b}} d x =\frac{\pi}{b} \csc \frac{(a+1) \pi}{b}\tag*{} $$ using my post. Differentiating $ J(a)$ w.r.t. $ a$ yields $\displaystyle \begin{aligned}I &=\frac{\partial}{\partial a} J(a) \\&=\frac{\partial}{\partial a}\left(\frac{\pi}{b} \csc \frac{(a+1) \pi}{b}\right) \\&=-\frac{\pi^{2}}{b^{2}} \csc \frac{(a+1) \pi}{b} \cot \frac{(a+1) \pi}{b}\end{aligned}\tag*{} $
Now I want to find the closed form of the integral in general
$$I_n=\int_{0}^{\infty} \frac{x^{a} \ln^n x}{1+x^{b}} d x, \tag*{} $$ where $b>a+1>0.$
$$\begin{aligned}I_n &=\frac{\partial^n}{\partial^n a} J(a) \\&=\frac{\pi^{n+1}}{b^{n+1}} \frac{d^n}{d x ^n}\csc x\Big|_{x=\frac{(a+1) \pi}{b}} \end{aligned} $$
Latest Edit
Noting that when $n$ is odd and $b=2(a+1)$,
\begin{aligned} I_{n} &=\int_{0}^{\infty} \frac{x^{a} \ln ^{n} x}{1+x^{2(a+1)}}dx \\ &=\left.\frac{\pi^{n+1}}{b^{n+1}} \frac{d^n}{d x ^n}\csc x\right|_{x=\frac{\pi}{2}} \\ &=0 \end{aligned}
we can deduce that
$$ \boxed{\int_{0}^{\infty} \frac{x^{a} \ln ^{2 m-1} x}{1+x^{2(a+1)}} d x=0} $$ for any natural number $m$ and $a>-1$.
My Question is how to find the last nth derivative. Opinions and solutions are warmly welcome.
There are several formulae for the $n^{\text{th}}$ derivative of $\csc (x)$ (have a look here for example). The simplest is probably $$\csc ^{(n)}(x)=i^{n+1} \left(\text{Li}_{-n}\left(-e^{i x}\right)-\text{Li}_{-n}\left(e^{i x}\right)\right)$$ Using it and trying to simplify as much as I could $$I_n=\int_{0}^{\infty} \frac{x^{a} \log^n( x)}{1+x^{b}} \,d x$$
$$\color{blue}{\frac{(2b)^{n+1}}{n!} I_n= }$$ $$\color{blue}{\Bigg[\zeta \left(n+1,\frac{b-(a+1)}{2 b}\right)-\zeta \left(n+1,\frac{2b-(a+1) }{2 b}\right)\Bigg]+}$$ $$\color{blue}{(-1)^n\Bigg[\zeta \left(n+1,\frac{(a+1)}{2 b}\right)-\zeta \left(n+1,\frac{b+(a+1)}{2 b}\right) \Bigg]}$$