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I've split the integral into:

$\int_0^1\frac{1-x^\alpha}{1-x}\mathrm dx=\int_0^{1/2}\frac{1-x^\alpha}{1-x}\mathrm dx+\int_{1/2}^1\frac{1-x^\alpha}{1-x}\mathrm dx$

I'm trying to find a suitable function in the form of $g(x)=\frac1{(x-1)^\text{?}}$ so I can use the LCT when $x\to 1$ but I can't figure out what needs to be the exponent. I need to somehow extract $(1-x)$ from $1-x^\alpha$.

How can this be done?

  • 0
    Maybe, if $\alpha$ is a positive integer you can use the fact that $\sum^{\alpha-1}_{k= 0} x^k= \frac{1-x^\alpha}{1-x}$?2011-02-09

3 Answers 3

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Evidently, $\alpha=0$ results in a convergent integral.

Suppose $\alpha>0$. Clearly, $\frac{1-x^{\alpha}}{1-x}$ is continuous on [0,1).

In fact, $\frac{1-x^{\alpha}}{1-x}$ is bounded on $[0,1)$ since $\displaystyle \lim_{x\to 1}\frac{1-x^{\alpha}}{1-x}=\lim_{x\to 1}\;\alpha x^{\alpha-1}=\alpha$.

It follows that the integral converges $\forall\alpha\geq0$.

  • 0
    $\frac{1-x^\alpha}{1-x}$ is unbounded near $0$ for \alpha < 0. Moreso, the integral does not converge for $\alpha \leq -1$. Take $\alpha = -1$ for example, the integrand simplifies to $\frac{-1}{x}$, which definitely does not converge near $0$.2011-02-10
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If $\alpha$ is a natural number, try factoring the numerator into something manageable. Notice that

$1-x^{\alpha} = (1-x)(1+x+...+x^{\alpha-2}+x^{\alpha-1})$

Now you can integrate each term.

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This is just an expansion on Shai Covo's comment above.

According to Gradshteyn & Ryzhik's Table of Integrals, Series, and Products, $\begin{align}\int_{0}^{1}\frac{1-x^\alpha}{1-x}dx &= \psi(\alpha + 1) - \psi(1)\\ &= \psi(\alpha + 1) + \gamma \end{align}$ and converges for $\text{Re} (\alpha) > 0$.

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    That's where analytic continuation comes in.2011-04-20