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Let $ f(x) = \frac{1}{x^n(1-x)} $ where $n\in \mathbb{N}$ and $x$ is not 0 or 1.

In a previous question, i determined that the partial fraction decomposition of $f(x)$ is

$ f(x) = \sum_{j=1}^{n}\frac{1}{x^j} + \frac{1}{(1-x)}$

Integrating this, when $n = 1$, I have:

$ \int f(x)\, dx = \ln|x|-\ln|1-x| + C = \ln\left|\frac{x}{1-x}\right|+C,$ for $x \neq 1,0$.

If $n >1$, I have (after some trial and error),

$ \int f(x)\, dx = \ln|x|+ \sum_{j=2}^{n}\frac{x^{-j+1}}{-(j-1)}-\ln|1-x| + C $

Have I missed anything here? Thanks for the feedback!

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    Sure, it's good. I guess I don't like negatives in denominator, so would prefer $-\sum_{j=2}^n \frac{x^{-j+1}}{j-1}$.2012-10-02

1 Answers 1

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For each $j\neq 1$, if

$g_j(x)=x^{-j}$

then

$\int g_j(x) dx =\frac{x^{1-j}}{1-j}+C$

Thus, given

$ f(x) = \sum_{j=1}^{n}\frac{1}{x^j} + \frac{1}{(1-x)}$

we have

$ \int f(x)dx = \int\frac1 xdx+ \sum_{j=2}^{n}\int \frac{dx}{x^j} +\int \frac{dx}{(1-x)}$

whence

$ \int f(x)dx = \log x+ \sum_{j=2}^{n}\frac{x^{1-j}}{1-j} +\log{(1-x)}+C$

You're right.

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    After looking carefully at your post (and admittedly not understanding every bit of it, as I studied similar material but in a completely different perspective), I see that your post emphasizes convergence. Some of your notion I don't fully follow, but I think I get the point. Thanks for sharing it with me. I had never heard of a majorant series.2019-04-03