$\int_{0}^{1} \frac{ \ln^n x}{1+x^{m}} d x$ =?

Question

Inspired by the post, I try to generalize the result to $$ I(m, n)=\int_{0}^{1} \frac{ \ln ^nx}{1+x^{m}} d x $$

Instead, I first investigate its partner integral $$ \begin{aligned} I(a): &=\int_{0}^{1} \frac{x^{a}}{1+x^{m}} d x . \\ &=\sum_{k=0}^{\infty}(-1)^{k} \int_{0}^{1} x^{a} \cdot x^{m k} d x \\ &=\sum_{k=0}^{\infty} \frac{(-1)^{k}}{a+m k+1} \end{aligned} $$

Differentiating $I(a)$ w.r.t. $a$ by $n$ times at $a=0$ yields $$ \begin{aligned} I(m, n)&=\left.\frac{{\partial}^{n}}{{\partial }a^{n}} I(a)\right|_{a=0}\\ &=\left.(-1)^{n} n ! \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(a+m k+1)^{n+1}}\right|_{a=0} \\ &=(-1)^{n} n ! \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(m k+1)^{n+1}} \\ &= (-1)^{n} n !\frac{(-1)^{n} n !}{m^{n+1}} \sum_{k=0}^{\infty} \frac{(-1)^{k}}{\left(k+\frac{1}{m}\right)^{n+1}} \\ &= \sum_{k=0}^{\infty} \frac{1}{\left(2 k+\frac{1}{m}\right)^{n+1}}-\sum_{k=0}^{\infty} \frac{1}{\left(2 k+1+\frac{1}{m}\right)^{n+1}}\\& = (-1)^{n} n !\frac{1}{2^{n+1}}\left[\sum_{k=0}^{\infty} \frac{1}{\left(k+\frac{1}{2 m}\right)^{n+1}}-\sum_{k=0}^{\infty} \frac{1}{\left(k+\frac{1}{2}\left(1+\frac{1}{m}\right)\right)^{n+1}}\right]\\ &=\frac{(-1)^{n} n !}{m^{n+1} 2^{n+1}}\left[ \zeta \left(n+1, \frac{1}{2 m}\right)-\zeta \left( n + 1 , \frac { 1 } { 2 } ( 1 + \frac { 1 } { m } ) \right)\right] \end{aligned} $$

My Question

I don’t know how to simplify the last answer. Would you help?

Answer 1

Take the $n$th derivative of both sides of

$$\int_0^1\frac{y^{s}}{1+y}\mathrm{d}y=\frac12\left[\psi\left(\frac{s+2}{2}\right)-\psi\left(\frac{s+1}{2}\right)\right],\quad \mathfrak{R}(s)>-1$$

w.r.t $s$, we get

$$\int_0^1\frac{y^{s}\ln^n(y)}{1+y}\mathrm{d}y=2^{-n-1}\left[\psi^{(n)}\left(\frac{s+2}{2}\right)-\psi^{(n)}\left(\frac{s+1}{2}\right)\right]$$

let $y=x^m$

$$\int_0^1\frac{x^{sm+m-1}\ln^n(x)}{1+x^m}\mathrm{d}x=(2m)^{-n-1}\left[\psi^{(n)}\left(\frac{s+2}{2}\right)-\psi^{(n)}\left(\frac{s+1}{2}\right)\right]$$

Finally set $s=\frac{1-m}{m}$

$$\int_0^1\frac{\ln^n(x)}{1+x^m}\mathrm{d}x=(2m)^{-n-1}\left[\psi^{(n)}\left(\frac{1+m}{2m}\right)-\psi^{(n)}\left(\frac{1}{2m}\right)\right],\quad n\in \mathbb{Z}^{+},\quad m>0.$$