# Integrating odd functions with infinite discontinuity:

If an odd function has an infinite discontinuity in its domain, can it be integrated (such that a convergent finite emerges) with that domain included?

For example: $$\int_{-1}^2 \frac{1}{x^{-3}} dx$$. Intuitively, it can be simplified to $$\int_1^2 \frac{1}{x^{-3}} dx$$ and thus the infinite discontinuity at 0 is removed.

If that is not doable, can an integral converge if the end points of the domain are infinite discontinuities?

For example: Does $$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} tan(x) dx = 0$$?

If these kinds of functions are split in two, and limits to $$\infty$$ are taken, then algebraic manipulation of infinities are required.

Any insight would be appreciated.

You should look into improper integrals?

Certainly, these are all improper integrals. But my question is whether the above integrals would converge or diverge. Intuitively/Geometrically, they should converge; but like I said, once the improper integral is carried out, one has to deal with infinities.

Gib Z
Homework Helper
Look up "Cauchy Principal Value". Strictly, the integrations over the discontinuities are divergent, but they can be assigned a Cauchy Principal Value that appeals to your geometric intuition.

Thanks a lot Gib Z! That helped immensely.