Integral of the day XIV

Integral involving the trigamma function $\psi^{(1)}(x)$

Today we show the proof of this result proposed by @Ali39342137 $$\int_{0}^{1} \frac{\ln(x)}{x^n+x^{n-1}+\cdots +1} dx = \frac{1}{(n+1)^2}\left[\psi^{(1)}\left(\frac{2}{n+1}\right)+ \psi^{(1)}\left(\frac{1}{n+1}\right)\right]$$

Proof:

Note that $$1+x+\cdots +x^{n-1}+x^n = \sum_{j=0}^{n} x^j = \frac{1-x^{n+1}}{1-x}$$ Hence $$I= \int_{0}^{1} \frac{\ln(x)}{x^n+x^{n-1}+\cdots +1} dx = \int_{0}^{1} \frac{(1-x)\ln(x)}{1-x^{n+1}} dx = \underbrace{\int_{0}^{1}\frac{\ln(x)}{1-x^{n+1}}dx}_{J} - \underbrace{\int_{0}^{1} \frac{x\ln(x)}{1-x^{n+1}}dx }_{K}$$ Since $0 \lt x \lt 1$ $$\begin{align*} J = \int_{0}^{1}\frac{\ln(x)}{1-x^{n+1}}dx =& \int_{0}^{1} \sum_{j=0}^{\infty} x^{(n+1)j} \ln(x) dx \\ =& \sum_{j=0}^{\infty}\int_{0}^{1} x^{(n+1)j} \left(\frac{d}{dt}\Big|_{t=0+} x^t \right) dx\\ =& \frac{d}{dt}\Big|_{t=0+} \sum_{j=0}^{\infty}\int_{0}^{1} x^{(n+1)j+t} dx\\ =& \frac{d}{dt}\Big|_{t=0+} \sum_{j=0}^{\infty}\frac{1}{(n+1)j+t+1}\\ =& \sum_{j=0}^{\infty}\left[\frac{d}{dt}\Big|_{t=0+}\frac{1}{(n+1)j+t+1}\right]\\ =& -\sum_{j=0}^{\infty}\frac{1}{\left((n+1)j+1\right)^2}\\ \end{align*}$$ $$\begin{align*} K= \int_{0}^{1} \frac{x\ln(x)}{1-x^{n+1}}dx =& \int_{0}^{1} \sum_{j=0}^{\infty}x^{(n+1)j+1}\ln(x)dx\\ =& \sum_{j=0}^{\infty}\int_{0}^{1}x^{(n+1)j+1}\left(\frac{d}{dt}\Big|_{t=0+} x^t \right)dx\\ =& \frac{d}{dt}\Big|_{t=0+} \sum_{j=0}^{\infty}\int_{0}^{1} x^{(n+1)j+1+t}dx\\ =& \frac{d}{dt}\Big|_{t=0+} \sum_{j=0}^{\infty}\frac{1}{(n+1)j+2+t} \\ =& \sum_{j=0}^{\infty}\left[\frac{d}{dt}\Big|_{t=0+}\frac{1}{(n+1)j+2+t} \right]\\ =& -\sum_{j=0}^{\infty}\frac{1}{\left((n+1)j+2\right)^2} \end{align*}$$ Hence $$\begin{align*} \int_{0}^{1} \frac{\ln(x)}{x^n+x^{n-1}+\cdots +1} dx =& -\sum_{j=0}^{\infty}\frac{1}{\left((n+1)j+1\right)^2} +\sum_{j=0}^{\infty}\frac{1}{\left((n+1)j+2\right)^2}\\ =& -\sum_{j=0}^{\infty}\frac{1}{(n+1)^2\left(j+\frac{1}{n+1}\right)^2} +\sum_{j=0}^{\infty}\frac{1}{(n+1)^2\left(j+\frac{2}{n+1}\right)^2} \end{align*}$$ Recall that he polygamma function has the series representation $$\psi^{(m)}(z) = (-1)^{m+1}\, m! \sum_{k=0}^\infty \frac{1}{(z+k)^{m+1}}$$ Therefore $$\begin{align*} \int_{0}^{1} \frac{\ln(x)}{x^n+x^{n-1}+\cdots +1} dx = & \sum_{j=0}^{\infty}\frac{1}{(n+1)^2\left(j+\frac{2}{n+1}\right)^2}-\sum_{j=0}^{\infty}\frac{1}{(n+1)^2\left(j+\frac{1}{n+1}\right)^2} \\ =& \frac{1}{(n+1)^2}\left[\psi^{(1)}\left(\frac{2}{n+1}\right)+ \psi^{(1)}\left(\frac{1}{n+1}\right)\right] \end{align*}$$ Hence, we can conclude $$\boxed{\int_{0}^{1} \frac{\ln(x)}{x^n+x^{n-1}+\cdots +1} dx = \frac{1}{(n+1)^2}\left[\psi^{(1)}\left(\frac{2}{n+1}\right)+ \psi^{(1)}\left(\frac{1}{n+1}\right)\right]}$$