I know that the length of the time interval of a function affects the accuracy of its Fourier Transform. So when we take a longer interval of a function, it is easier to distinguish between the different root frequencies.
I have already explained it this way in my paper. However, when applying to an example (transform of the same function, once over 5 seconds and the other time over 30 seconds):
$$ \hat{j}_5(f)=\int_{0}^{5} (\sin(4\pi t)+\sin(4.2\pi t))e^{-2\pi ift} \,dx=\frac{e^{-10\pi if}-1}{\pi f^2-4\pi}-\frac{1.05\left(e^{-10\pi if}+1\right)}{\pi f^2-4.41\pi} $$ $$ \hat{j}_{30}(f)=\int_{0}^{30} (\sin(4\pi t)+\sin(4.2\pi t))e^{-2\pi ift} \,dx=\frac{e^{-60\pi if}-1}{\pi f^2-4\pi}-\frac{1.05\left(e^{-60\pi if}+1\right)}{\pi f^2-4.41\pi} $$
I was interested to see that the only difference in the final equations was the exponential term. In $ \hat{j}_5(f) $ it is $ e^{-10\pi if} $ and in $ \hat{j}_{30}(f) $ it is $ e^{-60\pi if} $. The graph of $ \hat{j}_{30}(f) $ has a higher density of maxima and minima, and that appears to be the reason the frequencies 2Hz and 2.1Hz can be distinguished.
Values of $ \hat{j}_5(f) $
Values of $ \hat{j}_{30}(f) $
However, I want to explain this referring to the difference in exponential term. What is a valid explanation for it?