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How to compute


where $n>0$?

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Since $x^n \leq x$ in $[0,1]$, $$\int^1_0 \frac{1}{1-x^n} dx \geq \int^1_0 \frac{1}{1-x} dx$$ and the right integral is divergent. So your integral diverges.

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@BrianM.Scott I sure did! Feel free to edit such mistakes if you see me make them again =]. – Ragib Zaman May 19 '12 at 8:57
Okay; will do!$\;$ – Brian M. Scott May 19 '12 at 9:09
@RagibZaman I liked the first direction better. – Phira May 19 '12 at 9:51
@RagibZaman Maybe I am not clear enough: The current direction of the inequality is wrong. – Phira May 20 '12 at 15:17

Note that

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

for a function $f(x)$ that is continuous in the interval $[0,1]$, so the improper integral will exist for all $n \ge 1$ if and only if it exists for $n=1$.

But for $n=1$ you have an explicit antiderivative (using the logarithm), and you can check that it tends to infinity if you let the upper limit of the integral go to 1.

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