In the very last line of the solution that i have given i got the residue to be (e^(iaz))/2i which when multiplied by 2pii gives me pie^(iaz) now from this i don't understand how they got rid of z to write I(a) in terms of just a.
Any help would be much appreciated.
Assuming suitable conditions on $f$, $$ \widehat{f''}+2i\widehat{f'}-2\widehat{f} = \widehat{g} \\ -s^{2}\widehat{f}-2s\widehat{f}-2\widehat{f}=\widehat{g}\\ \{s^{2}+2s+1\}\widehat{f}=-\widehat{g} \\ \widehat{f}=-\frac{\widehat{g}}{s^{2}+2s+1} $$ Therefore, $$ f= -\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\frac{\hat{g}(s)e^{isx}}{s^{2}+2s+1}ds \\ = -\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}g(t)e^{-ist}\frac{e^{isx}}{s^{2}+2s+1}dtds \\ = -i\int_{-\infty}^{\infty}g(t)\frac{1}{2\pi i}\int_{-\infty}^{\infty}\frac{e^{is(x-t)}}{s^{2}+2s+1}ds dt \\ $$ The roots of $s^{2}+2s+1$ are $-1\pm i$. If $x-t > 0$, then $e^{is(x-t)}$ decays for $s$ in the upper half-plane, which allows you to close the contour in the upper half-plane and gives a value of $$ \frac{1}{2\pi i}\int_{-\infty}^{\infty}\frac{e^{is(x-t)}}{(s-(-1-i))(s-(-1+i))}ds \\ = \left.\frac{e^{is(x-t)}}{s-(-1-i)}\right|_{s=-1+i}=\frac{e^{-i(x-t)}e^{-(x-t)}}{2i} $$ If $x-t < 0$, then $e^{is(x-t)}$ decays for $s$ in the lower half-plane, which allows you to close the contour in the lower half-plane (but the contour is negatively oriented) and gives a value of $$ = -\left.\frac{e^{is(x-t)}}{s-(-1+i)}\right|_{s=-1-i}=-\frac{e^{-i(x-t)}e^{-(t-x)}}{-2i} $$ Combining these, $$ f(x) = \frac{-i}{2i}\int_{-\infty}^{\infty}g(t)e^{-i(x-t)}e^{-|x-t|}dt, $$ which can be written as $$ f(x) = -\frac{1}{2}\int_{-\infty}^{x}g(t)e^{-i(x-t)}e^{-(x-t)}dt -\frac{1}{2}\int_{x}^{\infty}g(t)e^{-i(x-t)}e^{-(t-x)}dt. $$ Now check the final answer: \begin{align} f'(x) = & -\frac{1}{2}g(x)-\frac{1}{2}\int_{-\infty}^{x}g(t)(-i-1)e^{-i(x-t)}e^{-(x-t)}dt \\ & +\frac{1}{2}g(x)-\frac{1}{2}\int_{x}^{\infty}g(t)(-i+1)e^{-i(x-t)}e^{-(t-x)}dt \\ f''(x) = & -\frac{1}{2}(-i-1)g(x)-\frac{1}{2}\int_{-\infty}^{x}g(t)(-i-1)^{2}e^{-i(x-t)}e^{-(x-t)}dt \\ & +\frac{1}{2}(-i+1)g(x)-\frac{1}{2}\int_{x}^{\infty}g(t)(-i+1)^{2}e^{-i(x-t)}e^{-(t-x)}dt \\ = & g(x)-\frac{1}{2}\int_{-\infty}^{x}g(t)(-i-1)^{2}e^{-i(x-t)}e^{-(x-t)}dt \\ & -\frac{1}{2}\int_{x}^{\infty}g(t)(-i+1)^{2}e^{-i(x-t)}e^{-(t-x)}dt \end{align} Therefore, $$ f''+2if'-2f= \\ = g(x) -\{(-i-1)^{2}+2i(-i-1)-2\}\frac{1}{2}\int_{-\infty}^{x}g(t)e^{-i(x-t)}e^{-(x-t)}dt \\ -\{(-i+1)^{2}+2i(-i+1)-2\}\frac{1}{2}\int_{x}^{\infty}g(t)e^{-i(x-t)}e^{-(t-x)}dt \\ = g(x). $$ I think there are errors in what was written in your problem. You can definitely see that what I have given you is correct because the final answer checks. So the final answer has been correctly verified to be $$ f(x) = -\frac{1}{2}\int_{-\infty}^{\infty}e^{-i(x-t)}e^{-|x-t|}g(t)dt. $$