# Why are some improper integrals convergent and others divergent?

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?

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}

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: 