How to check that $ f(x)=(1/2) \int_{-\infty}^{\infty} e^{-|x-y|} g(y) dy $ is a solution to $-f''+f=g$? where $f'',f',f \to 0$ as $ |x| \to \infty $ and $ g \to 0 \ $ as $|x| \to \infty $ ?
I do not understand how I would add $f''+f$ when one of them would be an integral and the other a function of $x$ ? I am also unsure of how I would get $f''$ since $f$ is a function of x and y. 
 A: With a fixed $x$, we can change limits of integral:
\begin{eqnarray}
2f(x)&=&\int_{-\infty}^{\infty}e^{-|x-y|}g(y)\,dy\\
&=&
\int_{-\infty}^{x}e^{-|x-y|}g(y)\,dy+\int_{x}^{\infty}e^{-|x-y|}g(y)\,dy
\\
&=&
\int_{-\infty}^{x}e^{-x+y}g(y)\,dy+\int_{x}^{\infty}e^{x-y}g(y)\,dy\\
&=&e^{-x}\int_{-\infty}^{x}e^{y}g(y)\,dy+e^{x}\int_{x}^{\infty}e^{-y}g(y)\,dy\\
&=& I+J
\end{eqnarray}
Where
$$I=e^{-x}\int_{-\infty}^{x}e^{y}g(y)\,dy~~~~,~~~~J=e^{x}\int_{x}^{\infty}e^{-y}g(y)\,dy$$
we know from $\displaystyle\lim_{x\to\pm\infty}g(x)=0$ that
\begin{eqnarray}
I'&=&-e^{-x}\int_{-\infty}^{x}e^{y}g(y)\,dy+e^{-x}\Big[e^{x}g(x)-\lim_{x\to-\infty}e^{x}g(x)\Big]\\
&=&-e^{-x}\int_{-\infty}^{x}e^{y}g(y)\,dy+g(x)\\
&=&-I+g(x)
\end{eqnarray}
\begin{eqnarray}
J'&=&e^{x}\int_{x}^{\infty}e^{-y}g(y)\,dy+e^{x}\Big[\lim_{x\to+\infty}e^{-x}g(x)-e^{-x}g(x)\Big]\\
&=&e^{x}\int_{x}^{\infty}e^{-y}g(y)\,dy-g(x)\\
&=&J-g(x)
\end{eqnarray}
S
\begin{eqnarray}
2f(x)&=&I+J\\
2f'(x)&=&-I+J\\
2f''(x)&=&I-g(x)+J-g(x)\\
&=&2f(x)-2g(x)\\
-f''(x)+f(x)&=&g(x)
\end{eqnarray}
as we desired.
