Prove that $\int_{-\infty}^{\infty} \sin x \, dx = 0 $

Question

$$\int_{-\infty}^{\infty} \sin x \, dx$$ When I am doing the proof for this, why do i have to split it into $\int_{-\infty}^a \sin x \, dx + \int_a^\infty \sin x \, dx $? where a is a constant

Answer 1

The assertion that the integral is $0$ doesn't really make sense: the convergence of this improper integral requires that both

\begin{gather} \lim_{a\to-\infty}\int_{a}^{0}\sin x \, dx \\ \lim_{b\to\infty}\int_{0}^{b}\sin x \, dx \end{gather}

exist and are finite and neither does. The "break point" $0$ is arbitrary and can be any real number. This is by definition of improper integral, at least the most common definition one finds.

The second limit doesn't exist, because if you compute it on the sequences $2n\pi$ or $2n\pi + \pi/2$ you get different limits:

$$ \lim_{n\to\infty}\int_{0}^{2n\pi}\sin x \, dx= \lim_{n\to\infty}[-\cos x]_{0}^{2n\pi}=0 $$ $$ \lim_{n\to\infty}\int_{0}^{2n\pi+\pi/2}\sin x \, dx= \lim_{n\to\infty}[-\cos x]_{0}^{2n\pi+\pi/2}=1 $$

In the same way you show that the first limit doesn't exist (just change the variable with $x=-y$).

Therefore, we can't say that $\displaystyle\int_{-\infty}^{\infty}\sin x\,dx$ is equal to a number, much less that it's zero, unless we give the symbol some other meaning than an improper integral.

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If you are considering the principal value, but you should clearly specify it, because it's a different thing than an improper integral, in general, you indeed get $0$, because the sine function is odd: $\sin(-x)=-\sin x$, so, for $a>0$,

$$ \int_{-a}^{a}\sin x\,dx = 0 $$

hence

$$ \mathrm{p.v.}\!\!\int_{-\infty}^{\infty}\sin x\,dx= \lim_{a\to\infty}\int_{-a}^{a}\sin x\,dx = 0 $$

The first equality in the above line is the definition of the principal value integral.

Answer 2

This is an improper integral, so, you have to define it as a limit. You have $$ \int_{-\infty}^\infty \sin(x)dx=\lim_{R\rightarrow\infty}\int_{-R}^R \sin(x)dx=0. $$ If you now take a different $a$, you have $$ \int_{-\infty}^a+\int_{a}^{\infty}\sin(x)dx=\lim_{R\rightarrow\infty}\int_{-R}^{-a}+\int_{-a}^{a}+\int_{a}^{R}\sin(x)dx=\int_{-a}^a\sin(x)dx=0 $$ and nothing changes. Thus, I think that you don't need to split it as $\int_{-\infty}^a+\int_{a}^{\infty}$.

PD: I consider the integral as Principal Value integral (when I pass to the limit the domain is symmetric). See the answer by @egreg