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!

You have done something strange with the lower limit. In fact, your integral is generalized at both endpoints, since the integrand is unbounded near $0$.
You get
$$\int_0^\infty \frac{1}{t^2}\,dt = \lim_{\varepsilon \to 0^+} \int_\varepsilon^1 \frac{1}{t^2}\,dt + \lim_{b \to\infty} \int_1^b \frac{1}{t^2}\,dt$$
and while the second limit exists, the first does not.
