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Is there a closed form (non-recursive) expression for the definite integral

$$\int_0^1 dz \frac{z^n}{z-a}, \qquad n\in\mathbb{Z}_+ \text{ and } a\notin (0,1)$$

for general $n$ and $a$ given in terms of logarithms? Mathematica is able to give them for any given integer $n$ (I made a table for $n=\{0,\ldots, 5\}$), but I can't find in the literature how the incomplete Beta function is related to logarithms:

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Anyone pointing me to the appropriate formulae in the NIST Handbook of Mathematical functions or in Gradshteyn and Ryzhik would be fantastic.

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  • $\begingroup$ Depends. What do you consider as a closed form? $\endgroup$ – Ali Caglayan Oct 19 '14 at 13:49
  • $\begingroup$ That's a good question, and something that I should have mentioned. The answer appears to be a sum of polynomial in $1/a$ and logarithm. So I guess I am looking for something of the form: $\sum_n c_n(1/a)^n + \text{logs}$. where the $c_n$'s are explicitly given (in terms of factorials, or Harmonic numbers, etc..) $\endgroup$ – QuantumDot Oct 19 '14 at 13:52
  • $\begingroup$ If you want it in terms of factorials then the Beta function can be written in terms of the Gamma function which can be written in terms of factorials thus a "closed" form. $\endgroup$ – Ali Caglayan Oct 19 '14 at 13:54
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Solution: You may write $$\begin{align} \int_0^1 \frac{z^n}{z-a}dz&=\int_0^1 \frac{z^n-a^n}{z-a}dz+a^n\int_0^1 \frac{1}{z-a}dz\\ &=\int_0^1 \sum_{k=0}^{n-1}a^{n-1-k}z^kdz+a^n\int_0^1 \frac{1}{z-a}dz\\ &=\sum_{k=0}^{n-1}a^{n-k-1}\int_0^1 z^kdz+a^n\left. \log (z-a)\right|_0^1\\ &=\sum_{k=1}^n\frac{a^{n-k}}{k}+a^n \log \left(1-\frac1a\right) \end{align} $$

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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{0}^{1}\dd z\,{z^{n} \over z - a}\,,\qquad n\in{\mathbb Z}_{+}\ \mbox{and}\ a \not\in \pars{0,1}:\ {\large ?}}$.

Lets $\ds{{\cal I}_{n} \equiv \int_{0}^{1}{z^{n} \over z - a}\,\dd z}$ such that

\begin{align} {\cal I}_{n}&=\int_{0}^{1}{z^{n - 1}\pars{z - a} + a z^{n- 1} \over z - a}\,\dd z ={1 \over n} + a{\cal I}_{n - 1} \end{align}

Then, \begin{align} {\cal I}_{n}&={1 \over n} + {a \over n - 1} + a^{2}\,{\cal I}_{n - 2} ={1 \over n} + {a \over n - 1} + {a^{2} \over n - 2} + a^{3}\,{\cal I}_{n - 3} \\[5mm]&=\cdots=\sum_{k\ =\ 0}^{n - 1}{a^{k} \over n - k} + a^{n}\,{\cal I}_{0} \quad\mbox{where}\quad {\cal I}_{0}=\int_{0}^{1}{\dd z \over z - a}=\ln\pars{\verts{1 - {1 \over a}}} \end{align}

$$\color{#66f}{\large% \int_{0}^{1}{z^{n} \over z - a}\,\dd z} =\color{#66f}{\large{\sum_{k\ =\ 0}^{n - 1}{a^{k} \over n - k} + a^{n}\,\ln\pars{\verts{1 - {1 \over a}}}}}\,,\qquad n\in{\mathbb Z}_{+}\ \mbox{and}\ a \not\in \pars{0,1} $$

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