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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.

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Now, to find $f(t)$, just take the inverse Fourier transform of $\frac{1}{i\omega+q}$ to get $ {{\rm e}^{-3\,t}}{\it H} \left( t \right)$, where $H(t)$ is the Heaviside function. –  Mhenni Benghorbal Mar 18 '13 at 20:02
    
@MhenniBenghorbal Could I then substitute this back in to show that it does satisfy the integral equation and if so how would I do this? –  Adam Mar 18 '13 at 23:10
    
Yes. It should satisfy the equation since it is a solution. –  Mhenni Benghorbal Mar 19 '13 at 14:59
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1 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 $$f(t)=e^{-3x}H(t).$$

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Could I then substitute this back in to show that it does satisfy the integral equation and if so how would I do this? –  Adam Mar 18 '13 at 23:11
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