# A rational function integral

How to evaluate : $$\int_{0}^{1}{\frac{{{x}^{a-1}}}{1+{{x}^{b}}}}\text{d}x,\ \ \ a,\ b\in {{\mathbb{N}}^{+}}$$

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$$\dfrac1{1+x^b} = \sum_{k=0}^{\infty}(-1)^k x^{bk}$$ Hence, $$I(a,b) = \int_0^1 \dfrac{x^{a-1}}{1+x^b} dx = \int_0^1 \sum_{k=0}^{\infty}(-1)^k x^{a+bk-1} dx = \sum_{k=0}^{\infty} \dfrac{(-1)^k}{a+bk} = \dfrac{\Phi(-1,1,a/b)}b$$ where $\Phi(z,s,a)$ is the Lerch transcendent function. If $a=b$, we get $$I(a,a) = \dfrac{\log 2}a$$

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I'm lazy and would just use a Taylor series since it's easy to compute from the geometric series formula.

\begin{align*}\int_0^1 \frac{x^{a-1}}{1+x^b} \, dx &= \int_0^1 x^{a-1} \left( \sum_{n=0}^\infty (-1)^n x^{bn} \right)\, dx \\&= \sum_{n=0}^\infty \int_0^1 (-1)^n x^{a-1+bn} \, dx \\&= \sum_{n=0}^\infty (-1)^n \frac{1}{a+bn} \end{align*}

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According to Maple: $${\mbox{_2F_1}(1,{\frac {a}{b}};\,{\frac {a+b}{b}};\,-1)}{a}^{-1}$$

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