How is $\int_{0}^{+\infty} x^{-2}dx$ divergent? I has just been introduced the definition of "improper integral" and proved that:   $$\int_{1}^{+\infty} x^{r}dx$$
converges if $r<-1$.
Then, why is it not so for $0$ to $+\infty$? Because we have
$$ \int_{0}^{x} t^{r}dt = \int_{0}^{1} t^{r}dt + \int_{1}^{x} t^{r}dt. $$
Therefore: $$\int_{0}^{+\infty} x^{r}dx = \lim_{x\rightarrow +\infty} \int_{0}^{x} t^{r}dt = \int_{0}^{1} t^{r}dt + \lim_{x\rightarrow +\infty}\int_{1}^{x} t^{r}dt$$ exists. So $\int_{0}^{+\infty} x^{r}dx$ converges??
 A: Assume $r\ne-1$.
One may observe that
$$
\left(\frac{t^{r+1}}{r+1} \right)'=t^r,\qquad t>0,
$$ giving, for each $a>0$,
$$
\int_a^1t^r\,dt=\frac{1}{r+1}-\frac{a^{r+1}}{r+1}.
$$
But if $r<-1$ (that is if $r+1<0$), we have
$$
\lim_{a \to 0^+}a^{r+1}=\infty
$$ and $\displaystyle\int_0^1\!t^r\,dt$ diverges.
A: This is one of the times that "area under the curve" can be a useful visualization.
Let's consider only cases where $r < 0,$ since I think you can easily figure out what happens if $r \geq 0.$
It is true that the integral
$$ \int_0^{+\infty} x^r \mathrm dx $$
converges if both of the integrals
$$ \int_0^1 x^r \mathrm dx 
\quad \text{and} \quad
 \int_1^{+\infty} x^r \mathrm dx $$
converge.
We can look at the integral $\int_1^{+\infty} x^r \mathrm dx$ as the area above the $x$ axis and under the curve $y = x^r$ to the right of the line $x = 1$.
We can likewise imagine that the integral $\int_0^1 x^r \mathrm dx$ is the area above the $x$ axis and under the curve $y = x^r$ between the lines $x=0$ and $x = 1$.
Part of that area is under the line $y = 1$ and won't give us any trouble;
it will always be a square of area $1$.
The other part is above the line $y = 1.$
We might notice that if we reflect the entire picture around the line $y = x,$
so that points that were on the $y$ axis change places with points that were on the $x$ axis, the area under $y = x^r$ and above $y = 1$ between $x=0$ and $x=1$
becomes the area under $x = y^r$ (that is, $y = x^{1/r}$) to the right of $x = 1.$
If that area is finite we can add $1$ to it (for the area of the square under the line $y=1$ between $x=0$ and $x=1$) and we will have the value of
$\int_0^1 x^r \mathrm dx$, which converges.
If that area is not finite then $\int_0^1 x^r \mathrm dx$ cannot converge.
What this suggests (I won't say "proves", since this is only a visualization)
is that the integral
$$ \int_0^1 x^r \mathrm dx $$
converges if and only if the integral
$$ \int_1^{+\infty} x^{1/r} \mathrm dx $$
converges.
And note that if $r \leq -1$ then $\frac 1r \geq -1$, so the second integral cannot converge in that case.
So we basically have a tug-of-war between the two parts of the integral,
the part from $x=0$ to $x=1$ and the part to the right of $x=1,$
to see who will make the entire integral from $0$ to $+\infty$ diverge.
If $-1 < r < 0$ then the area to the right of $x = 1$ is infinite
(or more formally, that part of the integral doesn't converge).
If $r < -1$ then the area above $y = 1$ is infinite, that is,
the part of the integral that goes from $0$ to $1$ doesn't converge.
If $r = -1$ exactly then both parts have infinite area and neither part of the integral converges.
