How to identify continuity or discontinuity of an [Definite] integral?

Question

How can I figure out whether an improper integral converges based on the discontinuities in the integrand?
For instance, these two both have discontinuities within the intervals of integration, and the second one also involves infinity: $$\int_{-4}^{4}\frac{1}{x}dx$$ what about this one: $$\int_{0}^{+\infty}\frac {ln(x)}{x}dx$$ Thanks

Answer 1

To determine whether the integral exists or not you need to treat them as limits around the discontinuities in the integrand.

The first one:

$$\int_{-4}^4{dx\over x}$$

exists iff both integrals:

$$\lim_{t\to 0}\int_{-4}^t{dx\over x},\quad \lim_{s\to 0}\int_s^4{dx\over x}$$

exist. But just check out the second one, the FTC gives this as:

$$\lim_{s\to 0}\log 4-\log s\to\infty$$

Similarly for the second one you need that both limits:

$$\lim_{t\to 0}\int_t^b{\ln x\over x}\,dx,\quad \lim_{s\to \infty}\int_b^s{\ln x\over x}\,dx$$

exist, but doing the first one and letting $u=\ln x\implies du={dx\over x}$ we are integrating

$$\lim_{t\to 0}\int_{\log t}^{\log b}u\,du ={1\over 2}\log^2 b-{1\over 2}\log^2t\to \infty$$

so it also diverges. You could as easily check the second integral diverges as well, you can make the same $u$-substitution and get

$$\lim_{s\to \infty}{1\over 2}\left(\log^2 s-\log^2 b\right)\to\infty$$

but only one of them needs to fail to exist to guarantee the whole integral fails to exist.

Just by way of contrast, let's see that it is possible that discontinuities don't stop us from integrating. Take

$$\int_0^1{dx\over\sqrt{x}}=\lim_{t\to 0}\int_t^1{dx\over x}$$

using the FTC we see this is:

$$\lim_{t\to 0}2\sqrt{1}-2\sqrt t=2$$