Why are some improper integrals convergent and others divergent?

Question

The integral of the function $f(x)=1/x^2$ is convergent and it equals 1 when the limits of the integral is $\int_1^\infty$ but it's divergent and equals $\infty$ when the limits are $\int_0^1$.
I know the math but I want to understand the reason intuitively(in layman's terms). Both of this function's parts look similar to each other(I know they are NOT identical) so Why don't they integrate to a similar value?.

Another example: the integral of the normal distribution is $1$ but the integral of the beta function(with $\alpha$ and $\beta$ equal $0$) $B(0,0)$ is $\infty$. Either x or y axis goes to infinity in both of those functions so Why are their integrals different?

Answer 1

Yes, they are similar. And both of them diverge. In fact\begin{align}\int_1^\infty\frac{\mathrm dx}x&=\lim_{M\to\infty}\int_1^M\frac{\mathrm dx}x\\&=\lim_{M\to\infty}\log M\\&=+\infty.\end{align}

Answer 2

You probably mean $f(x) = 1/x^2$ because

$$\int_0^1\frac{dx}x = \left[\begin{array}{cc}t=1/x &1\to 1\\ dx = -1/t^2\, dt &0\to \infty\end{array} \right] = \int_\infty^1-\frac{dt}{t} = \int_1^\infty\frac{dx}{x}.$$

In the case of $f(x) = 1/x^2$, we have $$\int_a^b\frac{dx}{x^2}= -\left.\frac 1x\right|_a^b=\frac 1a - \frac 1b,$$ so you can see that an improper integral converges as long as $0$ is not involved.

Intuitively, Riemann integral is about summing areas of infinitely many rectangles. If these areas decrease fast enough, integral converges and otherwise it doesn't.

Let me compare $\int_0^1 1/x^2\, dx$ and $\int_1^\infty 1/x^2\,dx$. Like before, consider the following

$$\int_0^1\frac{dx}{x^2} = \left[\begin{array}{cc}t=1/x &1\to 1\\ dx = -1/t^2\, dt &0\to \infty\end{array} \right] = \int_\infty^1 t^2\left (-\frac{dt}{t^2}\right) = \int_1^\infty 1\,dx.$$

The graph of $1/x^2$ approaches $x$-axis quickly enough as $x$ goes to infinity, so integral $\int_1^\infty 1/x^2\,dx$ converges, while the graph of $1$ doesn't approach $x$-axis at all, so $\int_0^1 1/x^2\, dx = \int_1^\infty 1\, dx$ diverges.

It might be more illustrating to compare $\int_1^\infty 1/x^2\,dx$ and $\int_1^\infty 1/x\,dx$. I'll just plot the graphs to show you how $1/x^2$ approaches $x$-axis much faster than $1/x$. Note that the axes ratio is $100:1$ to emphasize my point: