Solution of a simple integral equation

Question

I have the following integral equation: $$ f(x)=\exp{\left(-\int_{-x}^{\infty}f(y)\,dy\right)}\,\,, $$ where a condition on $f(x)$ holds: $$ f(x=0)=\frac{1}{2}\,. $$ I know that the solution is: $$ f(x)=\frac{1}{1+e^{x}} $$ but how can I show this (without knowing the solution)? I'm also interested in solving this equation using a software like Mathematica.

Answer 1

I came up with this solution thanks to one of the comments.

I differentiate both side: $$ \frac{d}{dx}f(x)=\frac{d}{dx}\exp{\left( -\int_{-x}^{\infty}f(y)\,dy\right)}\quad \Rightarrow \quad f'(x)=-f(-x)f(x)\,. $$ This equation implies that $f'(x)$ is an even function of $x$, which in turn implies: $$ f(-x)+c=-(f(x)+c)\quad \Rightarrow \quad f(-x)=-f(x)-2c\,, $$ where $c$ is a constant (that will be fixed imposing the last property at the end of the computation). Substituting this result in the differential equation for $f(x)$ I have $$ f'(x)=f(x)\big(2c+f(x)\big)\,. $$ This first-order non-linear differential equation can be solved by separation of variables: $$ \int \frac{1}{f(2c+f)}df=\int dx \quad \Rightarrow \quad \frac{1}{2c}\ln{\left(\frac{f}{2c+f}\right)} + K = x $$ where $K$ is a constant of integration that can be computed imposing that $f(0)=1/2$. With simple algebra it is possible to express $f$ as a function of $x$, depending also on the constants $c$ and $K'$ ($K'$ is another constant which depends on $K$ and can be found manipulating the previous equation for $f$): $$ f(x)=\frac{2c}{e^{-2cx}K'-1}\,. $$ Imposing $f(x=0)=1/2$, $K'$ can be fixed: $K'=4c+1$. At this point is possible to fix also $c$ imposing $f(-x)=-f(x)-2c$ for a simple value of $x$ (since this relation must hold for every value of $x\in\mathbb{R}$), for instance $x=0$, which leads to $c=-1/2$ and, finally: $$ f(x)=\frac{1}{1+e^{x}}\,. $$