How to show that $\int_0^\infty\frac{x}{1-x^2}dx$ indefinite?

Question

I'm thinking about the integral: $$\int_0^\infty\frac{x}{1-x^2}dx$$ The problem that this integral has is that it diverges at $x=1$. My question is now why one cannot think about the two diverging parts on the left and right of $x=1$ to cancel each other out. Is my understanding correct that this is basically the same problem as was asked here: Why does the integral of 1/x from negative infinity to infinity diverge? So that basically both parts give infinity and I can't just add negative and positive infinity to resolve the divergence of this integral? If this is too crude of an argument as to why the integral diverges then please let me know a cleaner argument/proof. Thank you in advance for your help!

Answer 1

You are probably thinking of Cauchy PV, which allows us to assign values to divergent integrals (as you say, “cancel out the infinities”). The definition of a Divergent Integral is:

The integral $\int \limits_a ^b f(x)\mathrm{d}x$ is called divergent if for some $ c \in (a,b)$ one of the integrals $\int \limits_a ^c f(x)\mathrm{d}x$, $\int \limits_c ^b f(x)\mathrm{d}x$ (or both of them) diverge.

In your case, take $ c=1$ to get that the integral from $0$ to $1$ diverges.