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We know that the Leibniz integral formula

$$\frac{d}{dt}\int_{\phi(t)}^{\psi(t)} f(t,s) ds = \int_{\phi(t)}^{\psi(t)} \frac{d}{dt}f(t,s) ds+f(t,\psi(t))\frac{d}{dt}\psi(t) -f(t,\phi(t))\frac{d}{dt}\phi(t).$$

Can we apply this rule for

$$\frac{d}{dt}\int_{\phi(t)}^{\infty} f(t,s) ds ?$$

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  • $\begingroup$ Yes under suitable decay assumptions. Let $c>0$ be fixed and apply the formula to $\int_{\phi(t)}^c$. Then let $c\to \infty$. Choose assumptions so that the $c$-dependent terms vanish in the limit. $\endgroup$ – Giuseppe Negro Mar 15 '15 at 23:14
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This form of Leibniz integral rule seems to require a interchange of limits. This is because, $$ \frac{d}{dt}\lim_{a\rightarrow\infty}\int_{\phi\left(t\right)}^{a}f\left(t,s\right)ds=\lim_{a\rightarrow\infty}\int_{\phi\left(t\right)}^{a}\frac{\partial}{\partial t}f\left(t,s\right)ds-f\left(t,\phi\left(t\right)\right)\phi'\left(t\right) $$ would follow immediately from the usual Leibniz integral rule so long as $$ \frac{d}{dt}\lim_{a\rightarrow\infty}\int_{\phi\left(t\right)}^{a}f\left(t,s\right)ds=\lim_{a\rightarrow\infty}\frac{d}{dt}\int_{\phi\left(t\right)}^{a}f\left(t,s\right)ds $$ But letting $$ g\left(t,a\right)=\int_{\phi\left(t\right)}^{a}f\left(t,s\right)ds $$ we can see that $$ \frac{d}{dt}\lim_{a\rightarrow\infty}g\left(t,a\right)=\lim_{a\rightarrow\infty}\frac{d}{dt}g\left(t,a\right) $$ requires some extra conditions. Namely, that $\frac{d}{dt}g\left(t,a\right)$ converges uniformly as $a\rightarrow\infty$ and that $\lim_{a\rightarrow\infty}g\left(t,a\right)$ converges for some t.

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$$\frac{\partial}{\partial t}\int_{\phi(t)}^{\infty} f(t,s)\ \partial s= \int_{\phi(t)}^{\infty} \frac{\partial}{\partial t} f(t,s)\ \partial s $$ $$=\lim\limits_{z\to \infty} \int_{\phi(t)}^{z} \frac{\partial}{\partial t} f(t,s)\ \partial s$$ $$ =\lim\limits_{z\to \infty}\left[f(t, z) \frac{\partial}{\partial t}z\right] - f(t, \phi(t)) \frac{\partial}{\partial t}\phi(t)$$ At this point, convergence is dependent on the term(s) in the limit.

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