Improper Integral - Odd Function and Symmetric range but Diverge

Question

I have learned the concept of improper integral and when it comes to the integral of an odd function, one needs to check the convergence of half interval first then if it converges, integral of total range would be $0$ e.g.

Here I have a question. Is there any function that it is odd, converges when it is integrated from $0$ to $\infty$ but diverges when it comes to the case of $-\infty$ to $+ \infty$?

Answer 1

No there is not.

To check if the integral of the function converges, you need to check that

$$\int_0^\infty \vert f\vert <+\infty.$$

Since $f$ is odd, if this value is indeed finite, you also have

$$\int_{-\infty}^0 \vert f\vert=\int_0^{+\infty} \vert f(-x)\vert\mathrm dx=\int_0^\infty \vert f\vert <+\infty.$$

So such a counter-example does not exist, since

$$\int_{-\infty}^{+\infty} \vert f\vert=2\int_0^{\infty} \vert f\vert <\infty.$$

Answer 2

No. You are assuming that $f$ is odd and that the limit $\lim_{R\to+\infty}\int_0^Rf(x)\,dx$ exists. But then$$\int_{-\infty}^{+\infty}f(x)\,dx=\lim_{R\to+\infty}\int_{-R}^Rf(x)\,dx=\lim_{R\to+\infty}0=0.$$The second equality comes from the fact that $f$ is odd.