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