About odd functions and improper integrals e.g. $\int^{\infty}_{-\infty}\sin x \; dx$

Question

Does $\displaystyle \int^{\infty}_{-\infty}\sin x \; dx$ converge?

Since $\sin x$ is an odd function, and we know that in definite integrals $\displaystyle \int^{a}_{-a}\sin x \; dx=0$ then does that simply implies to improper integrals as well?

Answer 1

No, in order for an integral to converge both $\int_0^{\infty} f(x) dx$ and $\int_{-\infty}^{0} f(x) dx$ must converge. $\int_{-\infty}^{\infty}\sin(x)dx$ converges in principal value however.

$\int_{-\infty}^{\infty}\sin(x)dx$ is DEFINED as the sum $\int_{-\infty}^{0} f(x) dx+\int_0^{\infty} f(x) dx$ where both converge, so no $\int_{-\infty}^{\infty} \sin(x) dx$ does NOT exist. However the way you are talking about is called the "Cauchy principal value" which means $\lim_{a\to\infty} \int_{-a}^{a} \sin(x) dx$ which DOES exist.