Definition of $\int_a^b f(x) \,\mathrm{d}x$ for $f$ continuous on $[ a,b )$ and unbounded at the right-hand endpoint How would I formulate a definition of the integral $\int_a^b f(x) \,\mathrm{d}x$ for a function continuous on $[a,b)$ and unbounded at the right-hand endpoint?
Could anyone provide an example?
This is the definition I have for a closed and bounded interval:

Let $f$ be a continuous function on an interval $[a,b]$ Then
  $$
\int\limits_a^b f(x) \,\mathrm{d}x = \lim\limits_{n\to\infty} \dfrac{b-a}{n}\sum\limits_{k=0}^{n-1} f \Bigg( a+\dfrac{k}{n}(b-a) \Bigg)
$$

I am thinking I need to use the left-hand endpoint to do this because the right-hand endpoint is unbounded. Thus the definition will have 
$$
\sum\limits_{k=0}^{n-1}f(\ldots)
$$
I am not really sure where to start though.
 A: The integral of Riemann doesn't exist for unbounded functions. This is the reason why mathematicians created the improper integral of Riemann.
The improper integral of Riemann is defined as a limit of proper integrals, in your case
$$\int_a^b f(x)\, dx:=\lim_{y\to b^-}\int_a^y f(x)\, dx$$

To show why the proper integral of Riemann doesn't exists for unbounded functions WLOG suppose that we have a function $f:[a,b]\to\Bbb R$ such that it is continuous to the left of some point $c\in[a,b]$ but $\lim_{x\to c^-}f(x)=\infty$.
Then we can choose a sequence of partitions $(P_k)$ of $[a,b]$ with decreasing mesh $\Delta_{P_k}=\delta_k$ for some sequence $(\delta_k)\downarrow 0$, such that in each $P_k$ there is an interval of the kind $(c-\delta_k,c]$. Then, because $f$ is continuous and unbounded to the left of $c$, for each $k$ there is some $x_k\in(c-\delta_k,c]$ such that $f(x_k)\ge k/\delta_k$.
Then if, for the interval $(c-\delta_k,c]$, we choose alternatively the tag $c$, when $k$ is odd, and the tag $x_k$ when $k$ is even, we find that the sequence of Riemann sums is not convergent, it will have two cluster points, hence the integral of Riemann is not defined for this function in $[a,b]$.

Note that the (proper) integral of Riemann is defined only in compact intervals, so an integral over a set of the kind $[a,b)$ is outside of the original definition of Riemann. To make sense we need to integrate on $[a,b]$ or $[a,c]$ for some $c\in[a,b)$.
If we try to apply the original definition of Riemann in an interval of the kind $[a,b)$ where $\lim_{x\to b^-}f(x)=\infty$ and $f$ is continuous in $[a,b)$ then we will get an integral that diverges to infinity.
A: Answer.
Definition Improper Riemann integral 
$$
\int_a^b f(x)\,dx=\lim_{\varepsilon\searrow 0}\int_a^{b-\varepsilon} f(x)\,dx
$$
