How to take this derivative

Question

My question is straightforward: I need to evaluate an expression of the form

$$ \frac{\partial}{\partial t}\sum_{k=0}^{t}\varphi(k,t) $$

How is this done, usually?

Answer 1

If instead of a sum you had an integral, things make a little more sense. Suppose $$ I(t)=\int_0^t\phi(s,t)\ ds. $$ To evaluate $I'(t)$, we could do the following: $$ \begin{align*} I(t)&=\int_0^{\infty}\chi_{[0,t]}(s)\phi(s,t)\ ds\\ &=\int_0^{\infty}\chi_{[s,\infty]}(t)\phi(s,t)\ ds\\ \end{align*} $$ Here $\chi$ denotes the characteristic function of a set. Assuming the integrand is sufficiently regular (i.e. a uniform $L^1$ bound local in $t$), we may differentiate under the integral sign to obtain $$ \begin{align*} I'(t)&=\int_0^{\infty}\frac{d}{dt}\left(\chi_{[s,\infty]}(t)\phi(s,t)\right)\ ds\\ &=\int_0^{\infty}\chi_{[s,\infty]}(t)\phi_t(s,t)+\delta(t-s)\phi(s,t)\ ds\\ &= \phi(t,t)+\int_0^t\phi_t(s,t)\ ds. \end{align*} $$

Answer 2

I found something that seems to work, actually, but that I have not rigorously proven does so.

Make the following construction:

$$ \phi(n) = \sum_{k=0}^n\varphi(n,k) = \int_0^{q_n}\varphi(n,k)dk $$ Then just take the forward difference ($\Delta_na(n) = a(n+1) - a(n)$) on both sides and solve the resulting recurrence for $q_n$: $$ \varphi(n+1,n+1) + \sum_{k=0}^n\Delta_n\varphi(n,k) = \int_0^{q_{n+1}}\varphi(n+1,k)dk - \int_0^{q_n}\varphi(n,k)dk;\hspace{5mm} \phi(n_0) = \int_0^{q_{n_0}}\varphi(n_0,k)dk $$ then just use the rules of differentiation under the integral as discussed in the original comments to get: $$ \frac{d\phi(n)}{dn} = \varphi(n,q_n)\frac{dq_n}{dn} + \int_0^{q_n}\frac{\partial\varphi(n,k)}{\partial n}dk $$

A simple example to try is $\varphi(n,k) = nk$, for which going through the above procedure yields $q_n = \sqrt{n(n+1)}$, and a quick check confirms that $$ \phi(n) = n\int_0^{\sqrt{n(n+1)}}k\hspace{0.7mm}dk = \frac{1}{2}n^2(n+1) $$ as we would expect. Not sure when this would fail, but the recurrences could get nasty very easily, so in any case might not always be the greatest path to take. Although for separable $\varphi$, $\varphi(n,k) = \nu(n)\mu(k)$, it's not super hard to show that the reucrrence reduces to a much nicer, $$ \mu(n+1) = \int_{q_n}^{q_{n+1}}\mu(k)dk. $$ and for which a (redundant) closed form solution is possible for $q_n$: $$ q_n = M^{-1}\left[\sum_{k=0}^{n-1}\left(\frac{dM}{dk}\right)_{k=k+1}\right] $$$$ M(k) \equiv \int\mu(k)dk $$