Can somebody please confirm or correct the following? If $a$ and $b$ are both positive integers such that $a<b$ and $b$ is even then we can write $$\frac{x^{a-1}}{x^{b}-1}=\frac{1}{b\left(x-1\right)}+\frac{\left(-1\right)^a}{b\left(x+1\right)}+\frac{2}{b}\sum\limits_{k=1}^{\frac{b-2}{2}}\frac{\cos\frac{2ak\pi}{b}\left(x-\cos\frac{2k\pi}{n}\right)-\sin^2\frac{2ak\pi}{b}}{x^2-2x\cos\frac{2k\pi}{b}+1},$$ and upon integration, I think we get $$\int\frac{x^{a-1}}{x^{b}-1}dx=\frac{1}{b}\log\left(x-1\right)+\frac{\left(-1\right)^a}{b}\log\left(x+1\right)+\frac{1}{b}\sum\limits_{k=1}^{\frac{b-2}{2}}\cos\left(\frac{2ak\pi}{b}\right)\log\left(x^2-2x\cos\frac{2k\pi}{b}+1\right)-\frac {2}{b}\sum\limits_{k=1}^{\frac{b-2}{2}}\frac{\sin^2\frac{2ak\pi}{b}}{\sin\frac{2k\pi}{b}}\tan^{-1}\frac{x-\cos\frac{2k\pi}{nb}}{\sin\frac{2k\pi}{b}}.$$

Context: I don't know my mistake at the moment, but I would like to get this instead $$\int\frac{x^{a-1}}{x^b-1}dx=\cdots-\frac{2}{b}\sum\limits_{k=1}^{\frac{b-2}{2}}\sin\left(\frac{2ak\pi}{b}\right)\tan^{-1}\frac{x-\cos\frac{2k\pi}{b}}{\sin\frac{2k\pi}{b}}.$$ That way, the formula could be used to derive another identity using the beta function which is my goal.

Sorry for length of the formula, I wasn't sure about appropriate question length.

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    $\begingroup$ Tiny error but I believe your first term in your integral should have a $\log(x-1)$ instead of $\log(x+1)$ and your second one the opposite. $\endgroup$ – Brad Mar 14 '14 at 22:43
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    $\begingroup$ Thank you, typo was corrected. I know what the problem is now, it was just a stupid mistake made at 3 am. $\endgroup$ – Hobbyist Mar 14 '14 at 23:03
  • $\begingroup$ As an aside, it may be more convenient to use partial fractions over the complexes rather than over the reals, so that all of your denominators are linear, rather than mostly quadratic. Also, it allows for the more compact notation of choosing a primitive $b$-th root of unity $\zeta$ and writing all of the constants in terms of powers of $\zeta$. $\endgroup$ – Hurkyl Mar 14 '14 at 23:12
  • $\begingroup$ Yes, the formula would be simpler to write in terms of $\zeta_b,$ I agree. But, that would not be useful when I take the formula and apply it to the identity I aimed to derive, which requires an expression in terms of sines and cosines. It's for this reason that I prefer quadratic factors here. Thanks for the help though. $\endgroup$ – Hobbyist Mar 14 '14 at 23:16

Mistake found

The rest of the partial fractions should have been found like so: $$\frac{e^{\frac{2ak\pi}{b}i}}{x-e^{\frac{2k\pi}{b}i}}+\frac{e^{-\frac{2ak\pi}{b}i}}{x-e^{\frac{-2k\pi}{b}i}}=\frac{2\cos\frac{2ak\pi}{b}\left(x-\cos\frac{2k\pi}{b}\right)-2\sin\frac{2ak\pi}{b}\sin\frac{2k\pi}{b}}{x^2-2x\cos\frac{2k\pi}{b}+1},$$ which corrects the original formula for the fraction to $$\frac{x^{a-1}}{x^{b}-1}=\frac{1}{b\left(x-1\right)}+\frac{\left(-1\right)^a}{b\left(x+1\right)}+\frac{2}{b}\sum\limits_{k=1}^{\frac{b-2}{2}}\frac{\cos\frac{2ak\pi}{b}\left(x-\cos\frac{2k\pi}{b}\right)-\sin\frac{2ak\pi}{b}\sin\frac{2k\pi}{b}}{x^2-2x\cos\frac{2k\pi}{b}+1},$$ which, upon integration, yields the formula $$\int\frac{x^{a-1}}{x^{b}-1}dx=\frac{1}{b}\log\left(x-1\right)+\frac{\left(-1\right)^a}{b}\log\left(x+1\right)+\frac{1}{b}\sum\limits_{k=1}^{\frac{b-2}{2}}\cos\left(\frac{2ak\pi}{b}\right)\log\left(x^2-2x\cos\frac{2k\pi}{b}+1\right)-\frac {2}{b}\sum\limits_{k=1}^{\frac{b-2}{2}}\sin\left(\frac{2ak\pi}{b}\right)\tan^{-1}\frac{x-\cos\frac{2k\pi}{b}}{\sin\frac{2k\pi}{b}}$$ which is the correct formula. The error was a simple error in expansion of the exponential terms.


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