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I have a function f defined by $f(x)=\frac{1}{1-x}$ on $[0,1)$ Is this function integrable and why.

I think it is integrable since the function is bounded and continuous on this interval. We could just assign a value to $f$ when x approaches 1.

Does this work well?

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The function is certainly not bounded on $[0,1)$. –  arkeet Dec 11 '12 at 7:56
    
Oh bother, just graphed it. Then I suppose the function is NOT integrable because it is not bounded or continuous on this interval. –  MathScratch Dec 11 '12 at 7:57
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up vote 1 down vote accepted

Note that for $x \in [0,1[$, $$ \frac 1{1-x} = \sum_{n=0}^{\infty} x^n, $$ hence when $x \to 1$, $\frac 1{1-x} \to \sum_{n=0}^{\infty} 1 = \infty$ (say by the monotone convergence theorem for instance). The reason why you can always integrate $f$ is because $f$ is positive on $[0,1[$. To see if $f$ is integrable (which is a different question), notice that $$ \int_0^1 \frac 1{1-x} \, dx = \int_0^1 \sum_{n=0}^{\infty} x^n \, dx = \sum_{n=0}^{\infty} \int_0^1 x^n \, dx = \sum_{n=0}^{\infty} \frac 1{n+1} = \infty. $$ where the swap of the series and the integral is given by the monotone convergence theorem of the Lebesgue integral.

If you use the Riemann integral, you can always compute $$ \int_0^1 \frac 1{1-x} dx = \lim_{a \nearrow 1} \int_0^a \frac 1{1-x} dx = \lim_{a \nearrow 1} -\log(1-a) = \infty. $$ EDIT : Andre Nicolas, I don't see why you deleted your answer. Your idea is perfectly fine : $$ \int_0^1 \frac 1{1-x} dx = \int_0^1 \frac 1u du. $$ (using the change of variables $1-x = u$). In this case we can only use the Riemann-integral trick though : $$ = \lim_{a \searrow 0} \int_a^1 \frac 1u du = \lim_{a \searrow 0} [\log(1) - \log(a)] = \infty. $$ If you are working with the Lebesgue integral, you can always truncate the function $\frac 1u$ and do something that allows you to get the equivalent of the limit using the monotone convergence theorem. But we're probably working too hard for this at this point.

Hope that helps,

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Thank you, but since the result is $\infy$ doesn't that mean that the function is not integrable? –  MathScratch Dec 11 '12 at 8:06
    
@MathScratch : Being able to integrate a positive function and it being integrable is two different things. Of course, in this case, $f$ is not integrable. For positive functions, even though the integral is defined, it does not mean that the function is integrable, because the latter asks for the integral (which is defined) to be finite. –  Patrick Da Silva Dec 11 '12 at 8:07
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