Derivative of a summation with variable in the upper limit I wish to derive the following equation with respect to its upper limit, but i'm not sure how to proceed.
$${ d \over dv} \sum_{t=1}^{T-t_0(v)} f(t)$$
This link offered some insight, as well as this link, altough i did not quite understand the solution of this second one. I thought of a solution like this.
Create a function using the following integral
$$
G(v,T,t) = 
\int_1^{T-t_0(v)} f(t) \, dt
$$
Then take the derivative of this function with respect to v:
$${ d \over dv} G(v,T,t)$$
This is correct?
 A: We can give a sound meaning to a sum with continuous (non-integral) bounds by the concept of Indefinite Summation.
This states that
$$
\eqalign{
  & f(x) = \Delta F(x) = F(x + 1) - F(x)\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad F(x) = \Delta ^{\left( { - 1} \right)} f(x) = \sum\nolimits_x {f(x)}
  = \sum\nolimits_{k = 0}^x {f(x)}  + c\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \sum\limits_{k = a}^b {f(k)}  = \sum\nolimits_{k = a}^{b + 1} {f(k)}
  = F(b + 1) - F(a) \cr} 
$$
which means that, if there is a  function $F(x)$ such that $F(x + 1) - F(x)=f(x)$, over $x \in \mathbb R$ or even $x \in \mathbb C$, then makes sense to put
that the sum of $f(x)$ between any real or complex bounds $a,b$ is $F(b)-F(a)$.
Which is a concept similar to the integral as anti-derivative.
Once you converted your sum into $F(b+1)-F(a)$, you can derivate wrt $b$ (or also wrt $a$).
Only to note that the traditional sum $\sum\limits_{k = a}^b$ converts in the new acception to $\sum\nolimits_{k = a}^{b + 1} $,
through the following steps
$$
\sum\limits_{k = a}^b {f(x)}  = \sum\limits_{a\, \le k\, \le \,b} {f(x)} 
 = \sum\limits_{a\, \le k\, < \,b + 1} {f(x)}  \to \sum\nolimits_{k = 0}^{b + 1} {f(x)} 
$$
Example:
$$
\eqalign{
  & \left( \matrix{  n + 1 \cr   3 \cr}  \right)
 = \left( \matrix{  n \cr   3 \cr}  \right) + \left( \matrix{  n \cr  2 \cr}  \right)
 \Rightarrow \Delta \left( \matrix{  n \cr  3 \cr}  \right)
 = \left( \matrix{  n + 1 \cr 3 \cr}  \right) - \left( \matrix{n \cr 3 \cr}  \right)
 = \left( \matrix{n \cr 2 \cr}  \right) = {{n\left( {n - 1} \right)} \over 2}  \cr 
  & \sum\limits_{k = 4}^7 {k\left( {k - 1} \right)}
  = \sum\nolimits_{k = 4}^{7 + 1} {k\left( {k - 1} \right)}
  = 2\left( {\left( \matrix{ 7 + 1 \cr 3 \cr}  \right) - \left( \matrix{ 4 \cr 3 \cr}  \right)} \right) = 104  \cr 
  & \sum\limits_{k = a}^b {k\left( {k - 1} \right)}
  = 2\left( {\left( \matrix{  b + 1 \cr 3 \cr}  \right) - \left( \matrix{ a \cr 3 \cr}  \right)} \right) \cr} 
$$
