Is the improper integral $\displaystyle \int_{-\infty}^{\infty} \sin (4x) dx$ convergent? $\bf{Try:} $
$\displaystyle \int_{-\infty}^{\infty} \sin (4x) dx=\displaystyle \int_{-\infty} ^{0} \sin (4x) dx +\displaystyle \int_{0}^{\infty} \sin (4x) dx $
Now let $\phi(t) =\displaystyle \int_{0}^{t}\sin (4x)dx \implies \phi(t) =\frac 14 - \frac{\cos 4t}4$
Now $\displaystyle \lim_{t\to \infty} \phi(t) =\displaystyle \lim_{n\to\infty} \left(\frac 14 - \frac{\cos 4t}4\right) $
This limit does not exist because $\cos$ is oscillating. Hence the improper integral is divergent.
But if we consider $\phi(t) =\displaystyle \int_{-t}^{t} \sin (4x)dx =0$ since $\sin$ is odd function. Here $\displaystyle \lim_{n\to \infty} \phi(t) =0\implies \text{the improper integral is convergent} $.
I'm confused which one to consider the first one or the second one? Please help me understand. Thanks in advance.
this two limits are in general different
$$ \lim_{(a,b)\rightarrow(-\infty,\infty)}\int^b_a f(x),dx $$ which is equivalent to $$ \lim_{a\rightarrow-\infty}\int^{x_0}a f(x),dx +\lim{b\rightarrow\infty}\int^b_{x_0}f(x),dx, $$ and $$ \lim_{b\rightarrow\infty}\int^b_{-b} f(x),dx $$
– Mittens Apr 10 '21 at 07:31