Let $f(x)$ be an integrable (not necessarily differentiable) function on $[a,\infty)$. Let $g(x,y)$ be differentiable on $[a,\infty)\times [c,d)$. I can also assume that the partial derivative of $g(x,y)$ with respect to $y$ is continuous.
Define $$h(y) = \int_a^\infty g(x,y)\,f(x)\,dx.$$
When calculating the differentiation $h'(y)$ for $y\in[c,d)$, can I always put the differentiation under the integral sign?
(If not, let me know if it will work with any additional "reasonable" assumptions.)
If $g_2(x,y)=\frac{\partial}{\partial y}g(x,y)$ is continuous in $y$, uniformly in $x$ (equicontinuous), then the Mean Value Threorem says $$ \begin{align} h(y+\Delta y)-h(y) &=\int_a^\infty(g(x,y+\Delta y)-g(x,y))f(x)\,\mathrm{d}x\\ &=\int_a^\infty\Delta y\,(g_2(x,y)+o(\Delta y))f(x)\,\mathrm{d}x\\ \end{align} $$ Therefore, $$ \begin{align} h'(y) &=\lim_{\Delta y\to0}\frac{h(y+\Delta y)-h(y)}{\Delta y}\\ &=\int_a^\infty g_2(x,y)f(x)\,\mathrm{d}x+\lim_{\Delta y\to0}\frac{o(\Delta y)}{\Delta y}\int_a^\infty f(x)\,\mathrm{d}x\\ &=\int_a^\infty\frac{\partial}{\partial y}g(x,y)f(x)\,\mathrm{d}x \end{align} $$