express integrals as limits How would you go about expressing the following as a limit?
$$\int_0^1 \ln(x) dx$$ 
I know how to express limits on simple equations, but have no clue how to go about expressing an integral as a limit.
Do I need to expand it as if I was performing a normal integration?
 A: This integral with unbounded integrand is improper at $x = 0,$ but it converges.
So
$$\int_0^1 \ln x \, dx = \lim_{c \to 0} \int_c^1 \ln x \, dx = \lim_{c \to 0}( x \ln x - x)|_c^1 = -1.$$
It also can be computed as the limit of a right-hand Riemann sum even though it is not properly Riemann integrable
$$\int_0^1 \ln x \, dx = \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \ln (k/n) = \ln \left(\lim_{n \to \infty} \left(\frac{n!}{n^n}\right)^{1/n}\right) = \ln(e^{-1}) = -1.$$
This is generally true for improper integrals with monotone integrands.
A: Since Riemann integration is only defined for bounded functions, technically the integral $$\int_0^1 \ln(x) dx$$ is not properly defined from the outset; we need to come up with a new definition for it. We see that $\ln(x)$ is bounded on any interval of the form $[\epsilon, 1]$ for $0 < \epsilon < 1$ so $\int_\epsilon^1 \ln(x) dx$ is properly defined for any such $\epsilon$. Hence, if the limit exists, we can define $$\int_0^1 \ln(x) dx := \lim_{\epsilon \to 0} \int_\epsilon^1 \ln(x) dx$$ and if the limit does not exist, we can declare the integral to be divergent (here you will find that that limit exists).
A: Integrate by parts
$$= \lim_{\epsilon \to 0^+} x \ln(x) |_\epsilon^1 - \int_0^1 1 \: \text{dx}$$
$$= -\left ( \lim_{\epsilon \to 0^+} \epsilon \ln(\epsilon) \right) - \int_0^1 1 \: \text{dx}$$
$$= -1$$
