Improper Riemann integral questions and determining whether they exist. Explain why
\begin{equation*}
\int_{0}^{1} \frac{dx}{x}, \quad \int_{1}^{\infty} \frac{dx}{x}
\end{equation*}
are improper Riemann integrals, and determine whether the limits they represent exist.
The improper Riemann integrals are the limits
\begin{equation*}
\begin{split}
\int_{0}^{1} \frac{dx}{x} &= \lim_{b\to 0^+} \int_{b}^{1} \frac{dx}{x} \\
&= \lim_{b\to 0^+} \left[\ln{(x)}\right]_{b}^{1} \\
&= \lim_{b\to 0^+} \left(-\ln{(b)}\right) \\
&= -\infty
\end{split}
\end{equation*}
and
\begin{equation*}
\begin{split}
\int_{1}^{\infty} \frac{dx}{x} &= \lim_{b\to \infty} \int_{1}^{b} \frac{dx}{x} \\
&= \lim_{b\to\infty} \left[\ln{(x)}\right]_{1}^{b} \\
&= \lim_{b\to\infty} \left(\ln{(b)}\right) \\
&= \infty.
\end{split}
\end{equation*}
So neither of these integrals exist.
My question is how do I explain why these are improper Riemann integrals? Thanks!
 A: Despite the imtegrand becoming $\infty$ at say $(x=a)$. One way it becomes improper but convergent is when $0<p<1$ where $$I=\int_{a}^{b} \frac{dx}{(x-a)^p}<\infty.$$ For instance
$$\int_{0}^{1} \frac{dx}{\sqrt{x}}=2$$ is improper but convergent. The integral $$\int_{0}^{1} \frac{dx}{x^{1.01}} =\infty,$$ is improper and divergent.
Next, one more way and integral becomes improper but convergent when $q>1$ where the integral is $$J=\int_{1}^{\infty} \frac{dx}{x^q}.$$ For instance $$\int_{1}^{\infty} \frac{dx}{x^{1.01}}=100, \int_{0}^{\infty} \frac{dx}{\sqrt{x}}=\infty$$
Finally both integrals of yours are inproper but divergent the first one is so as $p=1$ and second one is so as $q=1$.
Notice that the well known integral $$\pi=\int{-1}^{1} \frac{dx}{\sqrt{1-x^2}}=\int_{-1}^{1} \frac{dx}{(1-x)^{1/2} (1+x)^{1/2}},$$ is improper but convergent because for both $x=\pm 1$, $p=1/2<1.$
The integral $$\int_{0}^{1} \ln x~ dx =x\ln x-x|_{0}^{1}=-1.$$ is improper but convergent, In this case one has to take the limit $x\rightarrow 0$ of the anti-derivative.
The improper integrals do not connect well to area under the curve. This may
also be seen as a defect in the theory of integration or these integrals are  called improper which may or may not be convergent. 
There are many other ways and integral is improper but converget for instance
$$\int_{0}^{\infty} \frac{\sin x}{x} dx=\pi/2,$$ despite the integrand existing only as limit when $x \rightarrow 0.$
The integral $$\int_{0}^{\infty} \frac {dx}{(1+x^4)^{1/4}} \sim \int^{\infty} \frac{dx}{(x^4)^{1/4}} =\infty$$ is divergen, we know this without actually solving this integral!
Most often studying the integrand near the point of singularity (discontinuity) or $\infty$ and making out the value of $p$ or $q$ helps in finding if it is
convergent.
A: To note, there are two types of improper (Riemann) integrals. We consider the integral $\int_a^b f(x)dx$ for the purpose of description, where $a,b$ may be real numbers, $+\infty$, or $-\infty$.


*

*Type I: $f(x)$ is unbounded at some point in the interval of integration. That is, for some $c$ satisfying $a \le c \le b$, $\lim_\limits{x \to c} f(x) = +\infty$ (or $-\infty$).

*Type II: The interval itself is unbounded, i.e. at least one of $a,b$ are $+\infty$ or $-\infty$.
We now consider your integrals:
$$\int_0^1 \frac 1 x dx \;\;\;\;\; \int_1^\infty \frac 1 x dx$$

To see that the first is Type I, you simply need to recall the behavior of the function $f(x)=1/x$. It "blows up" towards infinity as $x$ approaches $0$ from the right - this is even more obvious if you graph the function. More symbolically, you could express this by saying $\lim_\limits{x \to 0^+} 1/x = +\infty$. 
We consider "from the right" here since $0$ is the lower bound of integration, and is less than the upper bound, $1$. You can easily imagine how this might extrapolate to the case where the upper bound is where it "blows up." For example, the below is also an improper Type I integral, but with the upper bound being the issue:
$$\int_0^1 \frac 1{x-1} dx$$
Note that Type I cases are not limited to the instances where the "blowing up" happens at the bounds, but also between the bounds. For example, this would also be an improper Type I integral:
$$\int_{-1}^1 \frac 1 x dx$$
Why? Recall: $\int_a^b f(x)dx + \int_b^c f(x)dx = \int_a^c f(x)dx$. By this property, you can split the above integral into two, one from $-1$ to $0$, and one from $0$ to $1$. Of course, both are Type I improper integrals.

Type II integrals, like your second, are easier to identify. They'll simply take the forms
$$\int_a^{\infty} f(x)dx \;\;\;\;\; \int_{-\infty}^b f(x)dx \;\;\;\;\; \int_{-\infty}^{\infty} f(x)dx$$
More simply, you can recognize Type II integrals by the fact that they have at least one bound at either $+\infty$ or $-\infty$. In that light it is immediately obvious why your integral is a Type II one.

Of course bear in mind that being Type I or Type II does not necessarily imply the integral itself does not exist. For example, it can be shown:
$$\int_0^1 \ln(x)dx = -1 \;\;\;\;\; \int_{-\infty}^\infty e^{-x^2} dx = \sqrt \pi$$
even though the first is Type I and the second is Type II. Not that noting this is strictly necessary but I feel it is worth noting. 
For the most part, your determination of the integrals existing - or, rather, not existing in this case - is fine. Just note you should have $+\infty$ for the first integral since $\lim_\limits{x \to 0^|} \ln(x) = -\infty$ (not positive). 
This is also evident from the "area under a curve" analogy for an integral: since $1/x > 0$ for all $x>0$, then the integral from $0$ to $1$ of it should be positive (you would have positive area if the function is positive everywhere there).
A: In the first case the limit from the right of the function at 0 goes to infinity (it does not exist). In the second case the interval of integration is unbounded.
