Is $\frac{1}{x}$ integrable? I am taking a course on theory of integrals and measure, where I have learned a very fundamental definition of integral. One of the things I've learned is that $f$ is integrable iff $\int|f|&lt;\infty$. Then, according to this definition, $f(x)=\frac{1}{x}$ wouldn't be integrable, since the integral of $|f(x)|=|\frac{1}{x}|$ diverges. However, I can calculate $\int_a^b\frac{1}{x}dx$ for any $a, b \in (-\infty, +\infty)$ and this isn't problematic at all. How do I deal with this?
I think your real difficulty is that we don't say &quot;$f$ is integrable,&quot; but rather &quot;$f$ is integrable over $D$&quot; where $D$ is some domain.
Your definition works fine if the domain is an interval that doesn't contain $0$.
Hint:
$$\int_a^b\frac{dx}x=\log\frac ba$$
(with no absolute value) is correct. This implies that $a,b$ must have the same sign and be nonzero. Otherwise the definite integral is undefined.