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Consider the function $$f(a) = \int^1_0 \frac {t-1}{t^a-1}dt$$ Can this function be expressed in terms of 'well-known' functions for integer values of $a$? I know that it can be relatively simply evaluated for specific values of $a$ as long as they are integers or simple fractions, but I am looking for a more general formula. I have found that it is quite easy to evaluate for some fractions--from plugging values into Wolfram I am fairly sure that $$f(1/n)=\sum^n_{k=1}\frac{n}{2n-k}$$ for all positive integer $n$. However, $f(a)$ gets quite complicated fairly quickly as you plug in increasing integer values of $a$, and I am stumped. Do any of you guys know if this function has a relation to any other special (or not-so-special) functions?

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3 Answers 3

up vote 3 down vote accepted

Notice that $$ \int^1_0 \frac {t-1}{t^a-1}dt= -\sum_{k=0}^{\infty}\int_{0}^{1} (t-1)t^{ka}dt = -\sum_{k=0}^{\infty}\frac{1}{(ka+1)(ka+2)}. $$

Now, if you sum the above series with help of maple or mathematica you get the same answer as in Jonathan's answer

$$ f(a)=\frac{1}{a}\Big(\psi\left(\tfrac{2}{a}\right)-\psi\left(\tfrac{1}{a}\right)\Big), $$

where $\psi(x)$ is the digamma function.

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Mathematica says that $$f(a)=\frac{1}{a}\Big(\psi\left(\tfrac{2}{a}\right)-\psi\left(\tfrac{1}{a}\right)\Big)$$ where $$\psi(z)=\frac{\Gamma'(z)}{\Gamma(z)}$$ is the digamma function.

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Let $\zeta$ be a primitive $a$-th root of unity. You can use a partial fractions expansion to rewrite

$$ \frac{t-1}{t^a - 1} = \sum_{i=0}^{a-1} \frac{\zeta^i - 1}{a \zeta^{i(a-1)} - 1} \frac{1}{t - \zeta^i} $$

and then integrate termwise. The result doesn't appear to simplify nicely, though.

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