(a) $\int_{-\infty}^{\infty} e^{-|t-x|}f(x) dx$ $= e^{-|t|} + |t|e^{-|t|}$.
It says to use the convolution theorem, which i understand to a certain extent. But I cant seem to think how to apply it/ solve this question. Any help would be greatly appreciated. I have my exam on this topic in 4 days.
Let be $$ \mathcal F\{f(t)\}=F(\omega)=\int_{-\infty}^\infty f(x) \,\mathrm e^{-i \omega t}\,\mathrm dt $$ the Fourier transform of $f(t)$. The Fourier transform of the convolution $$(f*g)(t)=\int_{-\infty}^\infty f(t) g(t-\tau)\,\mathrm d\tau =g(t)+|t|g(t)$$ where $g(t)=\mathrm e^{-|t|}$ is $$ \mathcal F\left\{(f*g)(t)\right\}=F(\omega)G(\omega)=G(\omega)+\mathcal F\left\{|t|g(t)\right\} $$ where $G(\omega)=\frac{2}{\omega^2+1}$ and the last Fourier transform is $\mathcal F\left\{|t|g(t)\right\}=2\frac{\omega^2-1}{(\omega^2+1)^2}$.
Thus we have $$ F(\omega)=1+\frac{\omega^2-1}{\omega^2+1}=\frac{2\omega^2}{\omega^2+1}=-(i\omega)^2G(\omega) $$ and finally $$ f(t)=-g''(t)=-g(t)+2\delta(t)=2\delta(t)-\mathrm e^{-|t|} $$