differentiation under integration of $e^{-yx}f(x)$ Let $f : (0,\infty) \rightarrow \mathbb{R}$ be a continuous function such that $\int_{(0, \infty)} |f| < \infty.$ Let $$F(y) = \int_0^\infty e^{-xy}f(x) dx$$ for all $y \in (0, \infty).$ Show that $$F'(y) = \int_0^\infty -xe^{-xy}f(x) dx.$$
It is a differentiation under integration sign. There is a theorem concerining it :
$\textbf{Theorem}$ For each fixed $y$, let $f(x,y)$ be an integrable function in $x$. Also, for each fixed $x$, assume that $\frac{d}{dy}f(x,y)$ exists. Then if there exists an integrable function $g(x)$ such that $$|\frac{d}{dy}f(x,y)| \leq g(x)$$ for all $x,y,$ $$\frac{d}{dy} \int_X f(x,y) dx = \int_X \frac{d}{dy}f(x,y) dx.$$
So in this problem, let $f(x,y) = e^{-xy}f(x)$ for $x,y \in (0,\infty).$ Then since $f \in L^1((0, \infty))$, $$\int_X f(x,y) dx \leq \int_X |f| < \infty.$$ So for each fixed $y$, $f \in L^1((0, \infty),x)$.
It is also clear that $\frac{d}{dy}f = -xe^{-xy}f(x)$ for all $x,y.$
So what left is the dominating function. Notice that $$|f'(x,y)| \leq xe^{-xy}|f(x)|.$$ (Remark : for short, $\frac{d}{dy}f(x,y) = f'(x,y)$) For big $y$, say $y \geq 1$, $$|f'(x,y)| \leq xe^{-x}|f(x)| \leq |f(x)|$$ which is integrable.
But for $y < 1$, I can just conclude that $$|f'(x,y)| \leq x|f(x)|$$ which is not guarantee to be integrable (I dont think it is true that $f \in L^1$ imply $xf(x) \in L^1$).
Any help for finding dominating functions ?
 A: Here's how I would approach this problem: in the theorem you mentioned, it should be the case (and you should check this with whatever source you obtained the theorem from) that we may conclude that the differentiation is valid at some $y_0$ if we may find $g$ satisfying the hypotheses on some range containing $y_0$. This makes sense because differentiation is a local property. Thus, we can simply show that for any $\epsilon>0$ that for all $y\in (\epsilon,\infty)$ that there is some integrable $g$ bounding $f'(x,y)$ on this range of $y$. The reason we have to do this is that $e^{-xy}$ "dampens" $xf(x)$ less and less as $y\rightarrow 0$ (and not at all at $y=0$),  but it works well enough if we restrict $y$ a set distance from $0$. What I mean is as follows:
$$|xe^{-xy}f(x)|\leq |{xe^{-x\epsilon}f(x)}|$$
for all $y\in(\epsilon,\infty)$. Now, it shouldn't be too hard to show that the function on the right-hand side is integrable since $xe^{-x\epsilon}$ is less than $1$ on $(M,\infty)$ for some $M$ and is clearly integrable on $(0,M)$. Since $\epsilon$ was arbitrary, we are done.
