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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!

Gerry Myerson
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JackReacher
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1 Answers1

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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.

Pedro
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    As a question of curiosity, when can we say $$\int \sum_{i}f_i(x) dx=\sum_i \int f_i(x)dx?$$ Is it always the case for indefinite integration? I recall having some information on this, but I've forgotten for some reason. – 000 Oct 02 '12 at 23:30
  • Disregard that; I found it: http://en.wikipedia.org/wiki/Fubini's_theorem – 000 Oct 02 '12 at 23:38
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    If the sum is finite, then you have no problem. For infinite sums, refer to this – Pedro Oct 02 '12 at 23:39
  • @Limitless How does that relate to your issue? – Pedro Oct 02 '12 at 23:40
  • I cannot articulate and portray a sincere understanding of the theorem (I've yet to cover all the concepts) in its entirety, but this answer seems to explain that: http://math.stackexchange.com/a/83747/22144

    The intuitive understanding I received from that essentially comes down to this: If the two different expressions are not divergent, then the swapping of the summation and integral sign is justified. I'll look more carefully at your answer (that you just linked) in just a moment. . .

     – 000 Oct 04 '12 at 00:27
  • 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. – 000 Oct 04 '12 at 00:36