I'm trying to solve the following integral equation using Fourier transforms: $$u(t)+ \int_{-\infty}^{t} e^{\tau-t} u(\tau)\,d\tau=e^{-2|\tau|}$$
I tried to transform both sides of the equation using $\mathfrak{F}\left[\int_{\infty}^{t} f(\tau)\,d\tau\right]=\frac{F(\omega)}{i\omega}+\pi F(0)\delta(\omega)$ which I found in a table of Fourier transforms but this did not seem to help me.
Is using Fourier transform a good approach for this equation? If it is, how do I proceed?
Let $$ \phi(t)=\begin{cases}e^t & \text{if }t\le0,\\0 & \text{if }t>0.\end{cases} $$ Then $$ \int_{-\infty}^{t} e^{\tau-t} u(\tau)\,d\tau=\int_{-\infty}^{\infty} \phi(\tau-t)\, u(\tau)\,d\tau=\phi\ast u(t). $$ Since the Fourier transform of a convolution is the product of the Fourier transforms, taking the Fourier transform on the equation we get $$ \hat u+\hat\phi\,\hat u=\widehat{\bigl(e^{-2|t|}\bigr)}. $$