Find the limit and integral $\lim_{\epsilon \to 0} \int_{\epsilon}^{1} \frac{x \sqrt{x} \log(x)}{x^4 + x^2 + 1} \, dx $

Question

Find the limit and integral$$ \lim_{\epsilon \to 0} \int_{\epsilon}^{1} \frac{x \sqrt{x} \log(x)}{x^4 + x^2 + 1} \, dx $$

My try

$$ \lim_{\epsilon \to 0} \int_{\epsilon}^{1} \frac{x \sqrt{x} \log(x)}{x^4 + x^2 + 1} \, dx = \int_{0}^{1} \frac{x \sqrt{x} (1 - x^2) \log(x)}{1 - x^6} \, dx $$

$$ = \int_{0}^{1} \frac{x \sqrt{x} (1 - x^2) \log(x)}{(1 - x^2)(1 + x^2 + x^4)} \, dx = \int_{0}^{1} \frac{x \sqrt{x} (1 - x^2) \log(x)}{1 - x^6} \, dx $$

$$ = \int_{0}^{1} \frac{x \sqrt{x} \log(x)}{1 - x^6} \, dx - \int_{0}^{1} \frac{x^3 \sqrt{x} \log(x)}{1 - x^6} \, dx = I_{1} - I_{2} $$

$$ I_{1} = \int_{0}^{1} \frac{x \sqrt{x} \log(x)}{1 - x^6} \, dx = \int_{0}^{1} x \sqrt{x} \sum_{k=0}^{\infty} x^{6k} \frac{\partial}{\partial a} \Big|_{a=0} x^a \, dx $$

$$I_{1} = \int_{0}^{1} \frac{x \sqrt{x} \log(x)}{1 - x^6} \, dx = \int_{0}^{1} x \sqrt{x} \sum_{k=0}^{\infty} x^{6k} \left. \frac{\partial}{\partial a} x^a \right|_{a=0} \, dx$$

Answer 1

Along the same lines as your attempt, we rewrite

$$I = \int_0^1 \frac{x \sqrt x \log x}{x^4 + x^2 + 1} \, dx \stackrel{x\mapsto x^4}= 4 \int_0^1 \frac{x^4 \log x}{x^8+x^4+1} \, dx = 4 \int_0^1 \frac{x^4 \left(1-x^4\right) \log x}{1-x^{12}} \, dx$$

Exploiting Taylor series leads to a result in terms of the polygamma function.

$$\begin{align*} I &= 4\sum_{n\ge0} \int_0^1 (x^{12n+4} - x^{12n+8}) \log x \, dx \\ &\!\!\stackrel{IBP}= 4\sum_{n\ge0} \frac1{(12n+9)^2} - \frac1{(12n+5)^2} \\ &= \frac1{36} \sum_{n\ge0} \frac1{\left(n+\frac34\right)^2} - \frac1{\left(n+\frac5{12}\right)^2} = \boxed{\frac{\psi\left(\frac34\right) - \psi\left(\frac 5{12}\right)}{36}} \end{align*}$$