Evaluating: $\int_0^{\infty} \frac{1}{t^2} dt$ I try to evaluate following integral $$\int_0^{\infty} \frac{1}{t^2} dt$$
At first it seems easy to me. I rewrite it as follows.$$\lim_{b \to \infty} \int_0^{b} \frac{1}{t^2} dt$$ and integrate the $\frac{1}{t^2}$. I proceed as follows:
$$\lim_{b\to\infty} \left[ 
-t^{-1}\right]_0^b$$ Then it results in,$$\lim_{b\to\infty} -b^{-1}$$ and it converges to zero. However, wolframalpha says the integral is divergent.
I do not understand what is wrong with this reasoning in particular and what is the right solution. Many thanks in advance!

The antiderivative is $-1/t$, which goes to $-\infty$ as $t \rightarrow 0$, so the integral is divergent.
