How to get h(t) using direct inverse Fourier transform formula for H(jw)=1/(a+jw) (only when |w|< W)?

I want to get the expression of signal in time domain by using inverse fourier transform. The signal in frequency domain is a little special: H(jw) is 1/(a+jw) only when |w|< W; and when |w|> W H(jw)=0. Can we use inverse fourier formula to get h(t)? In other words, can we get the expression of h(t) by calculating the integral below? $$h(t) = \frac{1}{2π}\int_{-W}^W \frac{e^{jwt}}{a+jw}dw\,.$$

• Assuming that $a$ is not purely imaginary, yes. You could also use some convolution result. – copper.hat May 20 '15 at 20:51
• I'm not sure there is an elementary answer. I thought your question was about existence. $\omega \mapsto {1 \over a+ i\omega}$ is the Fourier transform of $t \mapsto e^{-at} u(t)$, and $\omega \mapsto 1_{(-W,W)}(\omega)$ is the Fourier transform of a scaled version of $\text{sinc}$, so $h$ must be a convolution of these two. – copper.hat May 20 '15 at 21:36