Is this function integrable? 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?
 A: 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,
