Q) Use the residue theorem and contour integration to show that the inverse Fourier transform of the function

F[\omega ]=\frac{1}{\sqrt{2\pi }}\frac{1+j \omega }{1+\omega ^2}

is given by the function

f[x] = \left\{<br />
\begin{array}{cc}<br />
 e^x & x<0 \\<br />
 \frac{1}{2} & x=0 \\<br />
 0 & \text{otherwise}<br />
\end{array}<br />


My current strategy is to us the formula for the inverse Fourier transform

f[x]=\frac{1}{\sqrt{2\pi }}\int _{-\infty }^{\infty }F[\omega]e^{j \omega  x}dx

and integrate about a closed contour around the pole at \omega =-j, which would then equal 2\pi  j times the residue at that point ( e^x). This would then equal the integral along the real axis between -R and R, plus the integral along the open path given by the semi circle under the real axis which encircles the pole at -j. This all gives (I hope) the following

\underset{C}{\oint }\frac{e^{j \omega  x}}{\omega +j}d\omega =2\pi  j e^x=\underset{S}{\int }\frac{e^{j z x}}{z+j}dz-\int_{-R}^R \frac{e^{j \omega  x}}{\omega +j} \, d\omega

 \therefore  \int_{-R}^R \frac{e^{j \omega  x}}{\omega +j} \, d\omega =\underset{S}{\int }\frac{e^{j z x}}{z+j}dz - 2\pi  j e^x

My question is, am I going about this the right way? I've attempted to follow the method described above, but it leads to what seems to be a dead end (I get a horrific integral with exponentials of exponentials)...

Any help would be much appreciated!