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

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

is given by the function

$\displaystyle f[x] = \left\{

\begin{array}{cc}

e^x & x<0 \\

\frac{1}{2} & x=0 \\

0 & \text{otherwise}

\end{array}

\right.$

--------------------------------------------

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

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

and integrate about a closed contour around the pole at $\displaystyle \omega =-j$, which would then equal $\displaystyle 2\pi j $times the residue at that point ($\displaystyle 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

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

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

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!