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For improper integral $$\int_0^1 \frac{x^p}{1-x} \ln \left(\frac{1}{x}\right)\mathrm{d} x\qquad(p>-1),$$ we observe that there is a term $\frac{\ln(1)-\ln(x)}{1-x}$ in which we can use Taylor expansion of $\ln(1+x)$, then we can use $B$ function, but I'm stuck, I don't know how to simplify the result, the final result of this integral is $$\sum_{n=1}^{\infty }\frac{1}{\left ( n+p \right )^{2}}.$$ I don't know how to get to this point, could you please help me? Thank you.

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  • $\begingroup$ @Lorago it is dx $\endgroup$
    – 吴yuer
    Nov 30, 2023 at 12:46
  • $\begingroup$ Make the substitution $x=\exp(-t)$ and refer to $(25.11.25)$. $\endgroup$
    – Gary
    Nov 30, 2023 at 13:02

1 Answer 1

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$$\begin{align}\int_{0}^{1}\frac{x^{p}}{1-x}\ln\left(\frac{1}{x}\right)dx=&-\int_{0}^{1}\frac{x^{p}}{1-x}\ln\left(x\right)dx\\ =&-\frac{\partial}{\partial p}\int_{0}^{1}\frac{x^{p}}{1-x}\mathrm{d}x\\=&-\frac{\partial}{\partial p}\int_{0}^{1}\frac{x}{1-x} x^{p-1}\mathrm{d}x\\ \end{align}$$ $$\frac{x}{1-x}=\sum_{n=1}^{\infty}x^n$$ $$\begin{align}\int_{0}^{1}\frac{x^{p}}{1-x}\ln\left(\frac{1}{x}\right)dx=&-\frac{\partial}{\partial p}\int_{0}^{1}\sum_{n=1}^{\infty}x^n x^{p-1}\mathrm{d}x\\ =&-\frac{\partial}{\partial p}\sum_{n=1}^{\infty}\int_{0}^{1}x^{n+p-1}\mathrm{d}x\\ =&-\frac{\partial}{\partial p}\sum_{n=1}^{\infty}\frac{1}{n+p}\\ =&\sum_{n=1}^{\infty}\frac{1}{(n+p)^2}\\ =&\psi^{(1)}(p+1) \end{align}$$

Where $\psi^{(1)}(z)$ is the polygamma function

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