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And determine whether the limits they represent exist.

I evaluated both the integrals and showed that neither limits exist as finite numbers and so both integrals are divergent. I don't think I've answered the question correctly though, do I need to show that the function is Riemann integrable on both $(0, 1]$ and $[1, \infty)$?

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The Riemann integral is not defined on unbounded domains, such as $[1,\infty)$. This is why one of your integrals is improper. The Riemann integral is also not defined for unbounded functions. This is why your other integral is improper. –  David Mitra Jan 4 '13 at 3:24
    
@DavidMitra Would that be tantamount to saying that $\int_0^1 dx/x$ is an improper integral because $1/x$ is not continuous (not bounded) at 0 and that $\int_1^\infty dx/x$ is an improper integral since the upper limit of integration is infinite? Thanks for your help. –  user51462 Jan 4 '13 at 9:06
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@user51462: Not continuous and not bounded are different things, and it is unboundedness that is relevant here. For example, the function that is $0$ on the interval $[0,\frac12)$ and $1$ on the interval $[\frac12,1]$ is (properly) Riemann integrable on $[0,1]$, but it is not continuous. –  Jonas Meyer Jan 4 '13 at 9:09
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Let $f:[a,b] \to \mathbb{R}$ be some bounded function. Let $\mathcal{P} = \{a = x_0 < x_1 < ... < x_N = b\}$ be any partitioning of $[a,b]$. We define the (left) Riemann sum of $f$ corresponding to $\mathcal{P}$ by $\mathcal{R}_{\mathcal{L}}(f,\mathcal{P}) = \sum_{j=1}^N f(x_{j-1})(x_j - x_{j-1})$.

Define the mesh of $\mathcal{P}$ as $|\mathcal{P}| := \max \{x_j - x_{j-1}: 1 \le j \le N \}$. We say that $f$ is Riemann integrable if $\lim_{|\mathcal{P}| \to 0} \mathcal{R}_{\mathcal{L}}(f, \mathcal{P}) $ exists, in which case we denote this limit as $\int_a^b f(x) dx$.

Suppose now that we have a function $f:(a,b] \to \mathbb{R}$. For each $c >a$, we can ask whether $\int_c^b f(x) dx$ exists. If it does, and if $\lim_{c \to a^+} \int_c^b f(x) dx$ exists, we say that this value is an improper Riemann integral.

We similarly define $\int_a^{\infty} f(x) dx := \lim_{n \to \infty} \int_a^n f(x) dx$ and $\int_{-\infty}^b f(x) dx := \lim_{n \to -\infty} \int_n^b f(x) dx$.

In any case, the improperness only refers to the definition of the integral we're using, where we take a limit of well defined integrals to assign a new value to an integral that is not defined under our ordinary definition.

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