# What happens if we change the limits of integral in Fourier transform?

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?

• 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. – Dilip Sarwate Dec 15 '12 at 21:39