Evaluating: $\int_0^{\infty} \frac{1}{t^2} dt$

Question

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!

Answer 1

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.

Answer 2

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

Answer 3

Yes, it is divergent:

$$\int\limits_0^\infty\frac{dt}{t^2}=\left.\lim_{b\to\infty}\lim_{c\to 0^+}\left(-\frac{1}{t}\right)\right|_c^b=\lim_{c\to 0^+}\frac{1}{c}=\infty$$

Answer 4

The mistake you are making is that when you integrate you INCREASE the exponent of the variable by +1. By writing it as $\frac{t^{-3}}{-3}$ you are obviously REDUCING the exponent by 1. Hope it helps.

Answer 5

Note the lower bound $ x \to 0 , \frac 1 x \to \infty$, also you are subtracting power (on integration) instead of adding it.