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