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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?

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$\psi (\alpha + 1) = - \gamma + \int_0^1 {\frac{{1 - x^\alpha }}{{1 - x}}dx}$, where $\psi$ is the digamma function and $\gamma$ is Euler's constant. See en.wikipedia.org/wiki/Digamma_function. –  Shai Covo Feb 9 '11 at 17:48
    
Maybe, if $\alpha$ is a positive integer you can use the fact that $\sum^{\alpha-1}_{k= 0} x^k= \frac{1-x^\alpha}{1-x}$? –  Andy Feb 9 '11 at 19:38
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3 Answers 3

up vote 4 down vote accepted

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 $\forall\alpha\geq0$.

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i'm not sure i got why $\frac{1-x^{\alpha}}{1-x}$ is continuous on $[0,1]$. if $\alpha \lt 0$ then the function is not bounded near 0. it also has discontinuity at $x=1$. –  user6840 Feb 9 '11 at 21:24
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ok, i figured it out: since the function is bounded near 1, it isn't an improper point. when we deal with $$\int_{0}^{1/2}\frac{1-x^\alpha}{1-x}dx=\int_{0}^{1/2}\frac{1}{1-x}dx-\int_{0}^‌​{1/2}\frac{x^\alpha}{1-x}dx$$, the first term is a definite integral and the second we can compare to $g(x)=1/x^{-\alpha}$, then $$\lim_{x\to 0}=\frac{\frac{x^\alpha}{1-x}}{g(x)}=1$$ and by the LCT the integral converges iif $\alpha \lt 1$. –  user6840 Feb 9 '11 at 21:29
    
$\frac{1-x^\alpha}{1-x}$ 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 $0$. –  Brandon Carter Feb 10 '11 at 6:19
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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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FWIW: $\psi(\alpha)=-\gamma+\sum_{k=1}^{\alpha-1}\frac1{k}$ –  J. M. Apr 20 '11 at 14:19
    
@J.M.: How is the sum interpreted if $\alpha$ is not an integer? –  Brandon Carter Apr 20 '11 at 15:29
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That's where analytic continuation comes in. –  J. M. Apr 20 '11 at 15:32
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