I'm still rusty on integral equations. I need to solve the following
$$ f(x) = e^{-{\left | x \right |}} + \lambda \int_{-\infty}^{\infty} e^{-{\left | x - y \right |}} f(y)dy $$
where $f(x)$ is to remain finite for $x \rightarrow \pm \infty $
I'm guessing fourier transform?
On
While it is possible to solve using Fourier Transform, it is also possible to convert to a differential equation.
$$ f'(x) = -\frac{x}{|x|}e^{-|x|}+\lambda \int_{-\infty}^\infty -\frac{x-y}{|x-y|}e^{-|x-y|}f(y)dy $$ And, as $\frac{x}{|x|}=1$ when $x>0$ and $-1$ when $x<0$, we can write this as
$$ f'(x) = -\frac{x}{|x|}e^{-|x|}+\lambda\int_{-\infty}^x e^{x-y}f(y)dy-\lambda\int_x^\infty e^{y-x}f(y)dy $$ Note that we will only be considering $x\neq 0$ in terms of the differential equation - we will determine $x=0$ in a different manner. Differentiating a second time gives
$$\begin{align} f''(x) &= e^{-|x|}-\lambda f(x)+\lambda\int_{-\infty}^x e^{x-y}f(y)dy-\lambda f(x)+\lambda\int_x^\infty e^{y-x}f(y)dy\\ & = e^{-|x|}+\lambda\int_{-\infty}^\infty e^{-|x-y|}f(y)dy-2\lambda f(x)=(1-2\lambda)f(x) \end{align}$$ So, for $x\neq0$, the equation behaves as $f''(x)=(1-2\lambda)f(x)$. We will let $1-2\lambda=b^2$, to emulate the substitution used by Ron Gordon in his solution.
At $x=0$, we must look at the integral equation directly. That is,
$$ f(0) = 1+\lambda\int_{-\infty}^\infty e^{-|y|}f(y)dy $$ Now, solving our differential equation, we get
$$ f(x) = \left\{\begin{matrix} a_1 e^{bx}+a_2 e^{-bx}&\text{when }x>0\\f(0)&\text{when }x=0\\c_1e^{bx}+c_2e^{-bx}&\text{when }x<0\end{matrix}\right. $$ To keep the function bounded, we must have $a_1=c_2=0$. Evaluating the integral in our definition of $f(0)$, we get
$$ \int_{-\infty}^\infty e^{-|y|}f(y)dy = \int_{-\infty}^0 e^{y}\left(c_1e^{by}\right)dy+\int_0^\infty e^{-y}\left(a_2e^{-by}\right)dy $$ Given this, we have
$$\begin{align} \int_{-\infty}^\infty e^{-|y|}f(y)dy &= \left[\frac{c_1e^{(1+b)y}}{1+b}\right]_{-\infty}^0-\left[ \frac{a_2e^{-(1+b)y}}{1+b}\right]_0^\infty\\ &= \frac{a_2+c_1}{1+b} \end{align}$$ If we require continuity in the solution, then $$ a_2=1+\lambda\frac{a_2+c_1}{1+b}=c_1 $$ We can solve to get $a_2=c_1=\frac{1}{b}$
Substituting these into our function, we get $$ f(x) = \frac{1}{b}e^{-b|x|} = \frac{1}{\sqrt{1-2\lambda}}e^{-\sqrt{1-2\lambda}|x|} $$ Which can easily be recognised as the same result obtained by Ron Gordon.
Of course, the case of $\lambda=\frac12$ is a special case that must be handled separately. In this case, our differential equation is $f''(x)=0$, and so $f(x)=ax+c$, with $a$ and $c$ again not necessarily the same between $x<0$ and $x>0$. However, if $a\neq 0$, then $f(x)$ is not bounded, so, accounting for continuity, $f(x)$ must be a constant. Therefore, our $x=0$ condition says
$$ \int_{-\infty}^\infty e^{-|y|}f(y)dy = 2c $$ Therefore, we must have that $c=1+c$ which is a contradiction, and so there is no bounded solution for $\lambda=\frac12$.
You're guessing right. Use the convolution theorem and the fact that the FT of $e^{-|x|}$ is $2/(1+k^2)$. You will get an equation like
$$\lambda \frac{2 \hat{f}(k)}{1+k^2} + \frac{2}{1+k^2}=\hat{f}(k)$$
where $\hat{f}(k)$ is the FT of $f$.
Thus, the FT turns an integral equation into an algebraic one. Once you find $\hat{f}(k)$, take the inverse FT to solve for $f$.
$$\hat{f}(k) = \frac{2}{k^2+1-2 \lambda}$$
In order for the solution to be bounded, you should have $\lambda \ne 1/2$. In this case,
$$f(x) = \frac{1}{\pi} \int_{-\infty}^{\infty} dk \: \frac{e^{-i k x}}{k^2+1-2 \lambda} = \frac{1}{\sqrt{1-2 \lambda}} e^{-\sqrt{1-2 \lambda} |x|}$$
I have verified this solution to be correct by plugging it back into the equation and using the convolution theorem again; this involves evaluating the integral
$$\int_{-\infty}^{\infty} dk \: \frac{4}{(1+k^2)(b^2+k^2)} e^{-i k x}$$
where $b^2=1-2 \lambda$. The result is
$$\frac{2}{b (1-b^2)} e^{-b |x|} - \frac{2}{1-b^2} e^{-|x|}$$
which you can show that, when multiplied by $\lambda$ and added to the term $e^{-|x|}$, reproduces the solution.