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

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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).

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  • $\begingroup$ What does the Cauchy Principal Value mean in terms of areas? or what does it mean in general? $\endgroup$ – Jeel Shah Mar 24 '14 at 18:12
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    $\begingroup$ 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 $\endgroup$ – CameronJWhitehead Mar 24 '14 at 18:18
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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 $t$.

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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).

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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.

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