Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be continuous, bounded and consider $$ I(x) = \int_{-\infty}^{\infty} e^{-|y-x|} f(y) dy, \quad x \in \mathbb{R}. $$
What is the rigorous way of differentiating $I(x)$?
Some thoughts:
According to this book, if $( \Omega, \mathcal{F}, \mu)$ is a measurable space and $F : ( a, b ) \times \Omega \rightarrow \mathbb{R}$ is such that: $$ \int_{ \Omega } |F(x,y)| \mu (dy) < \infty \quad \forall x \in (a,b), \tag{1} $$
$$ x \mapsto F(x,y) \quad \text{is differentiable} \quad \forall y \in \Omega; \tag{2} $$
$$|\partial_x F(x,y)| \leq g(y) \quad \forall (x,y) \in (a,b) \times \Omega \quad \text{with} \quad \int_{\Omega}|g(y)| \mu(dy) < \infty; \tag{3}$$
then $$ \frac{d}{dx} I(x) = \frac{d}{dx} \int_{ \Omega } F(x,y) \mu(dy) = \int_{ \Omega } \frac{d}{dx} F(x,y) \mu(dy). $$
$(1)$ is fulfilled for $I(x)$. But for every fixed $y$ the function $x \mapsto e^{|y-x|} f(y)$ is not differenitable at $x = y$, we can therefore rewrite $$ I(x) = \int_{-\infty}^{\infty} e^{-|y-x|} f(y) dy = \int_{-\infty}^{x} e^{(y-x)} f(y) dy + \int_{x}^{\infty} e^{(x-y)} f(y) dy. $$ Now, for every $y$ the functions $x \mapsto e^{(y-x)} f(y)$ and $x \mapsto e^{(x-y)} f(y)$ are differentiable, but the limits of integration depend on $x$. Is it somehow possible to proceed? And what would be a proper choice of $g$? If there is some other result that can be applied, I would kindly ask for a reference.
In this particular case, you don't anything deep. You have $$ I(x) = \int_{-\infty}^{x} e^{(y-x)} f(y) dy + \int_{x}^{\infty} e^{(x-y)} f(y) dy=e^{-x}\int_{-\infty}^{x} e^{y} f(y) dy + e^x\int_{x}^{\infty} e^{-y} f(y) dy. $$ Now you can differentiate as a product and use the Fundamental Theorem of Calculus: \begin{align} I'(x)&=-e^{-x}\int_{-\infty}^{x} e^{y} f(y) dy+e^{-x}e^xf (x)+e^x\int_x^{\infty} e^{-y} f(y) dy-e^xe^{-x}f (x)\\ \ \\ &=-e^{-x}\int_{-\infty}^{x} e^{y} f(y) dy+e^x\int_x^{\infty} e^{-y} f(y) dy\\ \ \\ &=-\int_{-\infty}^{x} e^{y-x} f(y) dy+\int_x^{\infty} e^{-(y-x)} f(y) dy. \end{align} If you prefer to use the result you quoted, you can take $g (y)=e^{-|y|+c} $, with $c=\max (a,b) $.