Differentiating an improper integral depending on a parameter 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.
 A: 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) $.
A: $$I(x)=e^{-x}\int_{-\infty} ^{x} e^{y}f(y)dy+e^{x}\int_x^{\infty} e^{-y}f(y) dy.$$
Hence $$I'(x)=-e^{-x}\int_{-\infty} ^{x} e^{y}f(y)dy+f(x)+e^{x}\int_x^{\infty} e^{-y}f(y) dy-f(x)$$ $$=-e^{-x}\int_{-\infty} ^{x} e^{y}f(y)dy+e^{x}\int_x^{\infty} e^{-y}f(y) dy.$$
Differentiability of $f$ is not needed.
