How to get $h(t)$ using direct inverse Fourier transform formula for $H(jw)=1/(a+jw)$? I want to find the inverse Fourier transform of $H(jw)=1/(a+jw)$.
We know from the Fourier table that $$
F(e^{-at}) = 1/(a+jw).
$$ So that 
$$
h(t)=e^{-at}.
$$
But can we get $h(t)$ directly using inverse Fourier transform formula as below?
$$
h(t) = \int_0^\infty H(jw)e^{jwt}dw\,.
$$
Please help me. Thanks very much.
 A: Use contour integration.  
Case 1:  $t>0$
Observe the integral over $C$, where $C$ is comprised of the real line plus an infinite semi-circle, $C_{\infty}$, enclosing the upper-half plane of the complex $\omega $ plane. 
$$\frac{1}{2\pi}\oint_C\frac{e^{i\omega t}}{a+i\omega}d\omega=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{e^{i\omega t}}{a+i\omega}d\omega+\int_{C_{\infty}}\frac{e^{i\omega t}}{a+i\omega}d\omega$$
Use of Jordan's Lemma shows that the integration over $C_{\infty}$ vanishes.  Thus, we have
$$\frac{1}{2\pi}\oint_C\frac{e^{i\omega t}}{a+i\omega}d\omega=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{e^{i\omega t}}{a+i\omega}d\omega$$
The contour integral is equal to $2\pi i$ times the (only) residue at $\omega =ia$.  The residue is straightforward to compute and given by $-i e^{-at}$.
Thus, for $t>0$, 

$$H(\omega)=e^{-at}$$


Case 2:  $t<0$
Observe the integral over $C$, where $C$ is comprised of the real line plus an infinite semi-circle, $C_{\infty}$, enclosing the lower-half plane of the complex $\omega $ plane. 
$$\frac{1}{2\pi}\oint_C\frac{e^{i\omega t}}{a+i\omega}d\omega=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{e^{i\omega t}}{a+i\omega}d\omega+\int_{C_{\infty}}\frac{e^{i\omega t}}{a+i\omega}d\omega$$
Use of Jordan's Lemma shows that the integration over $C_{\infty}$ vanishes.  Thus, we have
$$\frac{1}{2\pi}\oint_C\frac{e^{i\omega t}}{a+i\omega}d\omega=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{e^{i\omega t}}{a+i\omega}d\omega$$
The contour integral is equal to $2\pi i$ times the residue in the lower-half plane.  However, there is no singularity in the lower-half plane and the residue is trivially zero.
Thus, for $t<0$, 

$$H(\omega)=0$$

