Express as a power series $\int_1^x \frac{\log t}{t-1} dt$ in $x-1$ and find the radius of convergence of the series

$$\int_1^x \frac{\log t}{t-1} dt$$

I assume that a power series in $x-1$ means having that form:$$ \sum_{n=0}^{\infty} a_n (x- 1)^n$$ My understanding is that I need to create a summation out of $\frac{\log t}{t-1}$ and then integrate term by term this summation.

I can see that $$\frac{\log t}{t-1} = - \frac{1}{1-t} \log t= \left( \sum_n^\infty t^n \right) \cdot (\log t)$$

But, I dont believe I can integrate term by term as $\log t$ is not yet part of the summation.

I attempted integration by part first in both configurations of $u$ and $dv$ with no satisfying results.

How would you do this? thanx for your input.


$$\int_{1}^{x}\frac{\log t}{t-1}\,dt\stackrel{t\mapsto s+1}{=}\int_{0}^{x-1}\frac{\log(s+1)}{s}\,ds=\int_{0}^{x-1}\sum_{n\geq 1}\frac{(-1)^{n+1}}{n}s^{n-1}\,ds $$ leads to the power series: $$ \sum_{n\geq 1}\frac{(-1)^{n+1}}{n^2}(x-1)^n =-\text{Li}_2(1-x)$$ whose radius of convergence is clearly one, the same as the radius of convergence of the Taylor series at the origin of $\frac{\log(s+1)}{s}$, or $\log(s+1)$, or $\frac{1}{s+1}$.

  • $\begingroup$ thx for the input. I understand that it is to "express the power series in $x-1$", is there any consideration to give to $t-1$ being replaced by $s$. If not What would be the justification? $\endgroup$ – rei Aug 16 '17 at 4:08

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