# Definite integral over singularity

It can be shown both heuristically and mathematically that the definite integral of an odd function over a symmetric interval is 0.

$$\int_{-a}^aOdd(x)dx=0$$

What I want to know is what is stopping me from evaluating

$$\int_{-a}^{a+\Delta}\frac{1}{x}dx$$

$$=\int_{-a}^{a}\frac{1}{x}dx+\int_{a}^{a+\Delta}\frac{1}{x}dx$$

$$=\int_{a}^{a+\Delta}\frac{1}{x}dx$$

Which for $a>0$ would return a finite number with no issue.

## 1 Answer

The problem is that $$\int \limits_{-a}^a \frac{1}{x} \, \mathrm{d} x = \int \limits_{-a}^0 \frac{1}{x} \, \mathrm{d} x + \int \limits_0^a \frac{1}{x} \, \mathrm{d} x = - \infty + \infty$$ is undefined, which reflects the fact the integral does not exist in the sense of the usual definitions of integrals (Riemann or Lebesgue).

However, intuitively we would like say that these infinities are equally large and should therefore cancel (just like the finite numbers would in the case of an odd function with a well-defined integral).

This intuition is captured by the definition of the Cauchy principal value of such an integral. It is indeed zero in our case, so we can write $$\mathrm{P}.\mathrm{V}. \int \limits_{-a}^{a+\Delta} \frac{1}{x} \, \mathrm{d} x = \int \limits_a^{a+\Delta} \frac{1}{x} \, \mathrm{d} x \, .$$