Is $\int_{-\infty}^{\infty} \sin x \, \mathrm{dx}$ divergent or convergent?

Question

I was determining whether

$$\int_{-\infty}^{\infty} \sin x \, \mathrm{dx}$$

was divergent or convergent. So, I did the following steps:

$$\begin{align} \int_{-\infty}^{\infty} \sin x \, \mathrm{dx} &= \int_{0}^{\infty}\sin x \, \mathrm{dx}+\int_{-\infty}^0\sin x \, \mathrm{dx} \\ &=\lim_{t\rightarrow\infty} \left(-\cos x |^{t}_{0}\right) + \lim_{a\rightarrow-\infty} \left(-\cos x |^{0}_{a}\right)\\ &=\lim_{t\rightarrow\infty} -\cos (t) + \cos 0 + \lim_{a\rightarrow-\infty} -\cos 0 + \cos a\\ &=\lim_{t\rightarrow\infty}1 - \cos t + \lim_{a\rightarrow-\infty} -1+\cos a \end{align}$$

Now, at this point, it would be reasonable to say that both the limits are undefined and therefore, the integral is divergent but then if I try something like the following

\begin{align} \quad\quad&=\lim_{t\rightarrow\infty}1 -\cos t + \lim_{a\rightarrow\infty} -1+\cos a \\ \quad\quad&=\lim_{t\rightarrow\infty}-1 -\cos t + \lim_{a\rightarrow\infty} \cos a+1 \\ \quad\quad&=\lim_{b\rightarrow\infty}-1 +\cos b - \cos b+1 \\ \quad\quad&= 0 \end{align}

So, as you can see, it was shown before that the integral is divergent but with some manipulation, we came at an answer of $0$ but is that valid? I assume, a similar technique can be applied to $\int_{-\infty}^{\infty} \frac{1}{x} \, \mathrm{dx}$.

Answer 1

Your first claim was correct: the limit does not exist. $t$ and $a$ are unrelated, so there's no good reason you should be able to set $t=a=b$ and take a limit. For $\int_{-\infty}^\infty \sin x dx$ to be defined, both $\int_{-\infty}^0 \sin x dx$ and $\int_{0}^{\infty} \sin x dx$ must exist: but as you saw, neither do.

What you calculated is instead called the Cauchy Principal Value; indeed, the Cauchy principal value of $$\int_{-\infty}^\infty \sin(x) dx$$ is $0$ (as it is for every odd function).

Answer 2

The problem with $\int_0^t\sin(x)dx $ is that this (as a function of $t$) oscillates around $0$. With each period of the integrand you first add, then remove the same amount indefinitely. Therefore it does not converge.

$\int \frac{1}{x}$ is different. Both $\int_0^1 \frac{1}{x}$ and $\int_1^\infty \frac{1}{x}$ diverge, without oscillating. The finite integral $\int_1^t \frac{1}{x}$ is, e.g, positive for all $t$.

Answer 3

Maybe it is helpful.

The inverse Fourier transformation of the Dirac-delta function is $$ \delta (t) = \frac{1}{2\pi} \int_{-\infty}^\infty \mathrm{e}^{\mathrm{i} \omega t} \mathrm{d} \omega $$

By letting $\omega = x$, it is shown that $$ \int_{-\infty }^\infty \sin x\mathrm{d} x= \Im(2\pi \delta (t=1)) = 0 $$

Answer 4

In Fourier analysis (particularly in physics), a technique of adding $e^{-\lambda |x|}$ into the integral, integrating and limiting $\lambda \to 0$ is commonly used as a regularization method. But otherwise, as others pointed out, the integral diverges, because no matter how far you integrate, the partial result doesn't converge (doesn't stop chaging).

Answer 5

The second way you use to compute the integral is not legit, since the limits in the improper integral need to be computed separately, in general. Hence, the integral is divergent.