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

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

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 $\int_{-\infty}^{\infty} \frac{1}{x} \, \mathrm{dx}$.

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).

• What does the Cauchy Principal Value mean in terms of areas? or what does it mean in general? – Jeel Shah Mar 24 '14 at 18:12
• The Cauchy Principal Value is $\lim_{n \to \infty} \int_{-n}^{n} f(x) dx$ which is $0$ in this case (and yeah, for any odd function this is zero). The important thing (which is what makes this not integrable) is that $\lim_{n \to \infty} \int_{-n}^{2n} f(x) dx$ is not $0$. If we say a function is integrable on a set we want the integral to be independent of the way we "take the integral". This is analogous to when a sum is absolutely convergent we can rearrange the terms without changing the value of the sum, but when we change the order of terms in a conditionally convergent sum it may change – CameronJWhitehead Mar 24 '14 at 18:18

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 $t$.

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 $$\int_{-\infty }^\infty \sin x\mathrm{d} x= \Im(2\pi \delta (t=1)) = 0$$
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).