When we take the Fourier transform of $\cos({2 \pi f_0 x})$ on a finite interval $[-a, a]$, even though we increase the $a$ too much, there is still contributions from other frequencies rather than $f_0$. Why don't the contributions other frequencies in the Fourier transform graph become zero as we increase the integral range?
$$ \int_{-a}^{a}{\cos\left(2{\pi}f_0x\right)\cdot\mathrm{e}^{-2\mathrm{i}{\pi}fx}} = \dfrac{\left(f_0-f\right)\sin\left(2{\pi}af_0+2{\pi}af\right)+\left(f_0+f\right)\sin\left(2{\pi}af_0-2{\pi}af\right)}{2{\pi}\cdot\left(f_0^2-f^2\right)} $$
For $f_0 = 3$;
$$
\dfrac{\left(f-3\right)\sin\left(2{\pi}af+6{\pi}a\right)+\left(f+3\right)\sin\left(2{\pi}af-6{\pi}a\right)}{2{\pi}\cdot\left(f^2-9\right)}
$$
When we take the integral over the $[-4,4]$, the Fourier transform looks as following.
over the $[-8,8]$,
over the $[-40,40]$
Why don't the contributions from other frequencies become close to zero as we increase the range to $[-\infty, \infty]$? How does one explain this?