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Why can't I apply Leibniz' rule in the following way? $$\frac{d}{ds} g(s)\int_0^\infty f(s,x,u) \, du = \int_0^\infty \frac{d}{ds}g(s)f(s,x,u)\,du,$$ assuming $gf$ and $(gf)'$ are continuous on $[0,+\infty]\times [s_0,s_1]$ for some $s_0<s_1\in\mathbb{R}$. |
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The problem I was originally having was in thinking that $g(s)$ was a constant coefficient of the definite integral. However, since we wish to differentiate w.r.t. $s$ it seems we can not think of $g$ as a constant, which seems obvious now. Instead, we must treat $g\int$ as a product and apply the product rule of differentiation, e.g. $$\frac{\partial}{\partial s}g(s)\int_0^\infty f(s,x,u)du = g(s)\frac{\partial}{\partial s}\int_0^\infty f(s,x,u)du + \left(\frac{d}{d s}g(s)\right)\int_0^\infty f(s,x,u)du.$$ We can then take the $\partial/\partial s$ inside the integral as required. |
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