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By definition of Fourier transform

$$X(\omega)=\int_{-\infty}^\infty x(t) e^{-j\omega t} dt $$

Now what will happen to the answer of transform for example in case of $x(t)= \cos(\omega_0 t)$ if limit is $0$ to $A$ instead of $-\infty$ to $\infty$?

For $x(t)=\cos(\omega_0 t)$ its fourier transform is given by $ X(\omega)= \pi[\delta(\omega-\omega_0) + \delta(\omega+\omega_o)]$

so if the limit is changed will it effect the answer?

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Yes, it will affect the answer. What you're suggesting is known as the short-time Fourier transform. In the sinusoidal case that you proposed, you will observe spectral leakage, as the truncation of the integral limits is equivalent to multiplication of the sinusoid by a rectangular window function. This multiplication in the time domain maps to convolution in the frequency domain. The Fourier transform of a rectangular window is a sinc function, so the convolution will yield two sinc functions centered at the locations of the impulses in your original answer.

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    $\begingroup$ Moreover, with limits $0$ and $A$, the Fourier transform is definitely going to be a complex-valued function of $\omega$ while if the limits were $-A$ and $A$, the Fourier transform will continue to be purely real-valued. $\endgroup$ – Dilip Sarwate Dec 15 '12 at 21:39

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