The function $f(t)$ satisfies the integral equation
$f(t)+2\int_{-\infty}^{\infty}H(s)e^{-s}f(t-s)ds=H(t)e^{-t}$ and decays as t $\rightarrow_{-\infty}^{\infty}$
By taking the Fourier transform of the equation with respect to t, show that $\tilde{f}(\omega) = \frac{1}{3+\omega}$
Hence obtain an expression for $f(t)$
I have previously computed the Fourier transform of the function $h_{q}(t)=H(t)e^{-qt}$
where $q>0$ is a real constant and $H(t)$ is the Heaviside step function defined by
$H(t) = 0: t<0$
$H(t)=\frac{1}{2} : t=0$
$H(t)= 1 : t>0$
This came to $\frac{1}{i\omega+q}$
I know I have to use this result and the convolution theorem now but I'm not sure how to go about it.
Best Answer
We denote by $g(t)=H(t)e^{-t}$ then $$\mathcal{F}(g)(w)=\frac{1}{iw+1},$$ and we have $$f+2g*f=g$$ and by Fourier transform we find $$\mathcal{F}(f)(w)=\frac{\mathcal{F}(g)(w)}{2\mathcal{F}(g)(w)+1}=\frac{1}{iw+3}.$$
Now we know that (see Fourier transform) $$\mathcal{F}(e^{-ax}H(t))(w)=\frac{1}{a+iw}$$ where $H(t)$ is the Heaviside unit step function and $a>0$, then we have $$f(t)=e^{-3x}H(t).$$