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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?

MrCorote
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  • The summation and integral are only equal for certain functions, do you have any restrictions on $f$ and $t_0$ – Henry Lee Feb 10 '21 at 20:05
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    I wished to make an easier illustration of the problem. The function $f(t) = e^{-pt}$ and the function $t_0(v) = e^{-v^{\beta}}$, with $\rho > 0$ and $\beta >0$. Also, $v$ is stritctly postive. – MrCorote Feb 10 '21 at 21:10

1 Answers1

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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: $$ \eqalign{ & \left( \matrix{ n + 1 \cr 3 \cr} \right) = \left( \matrix{ n \cr 3 \cr} \right) + \left( \matrix{ n \cr 2 \cr} \right) \Rightarrow \Delta \left( \matrix{ n \cr 3 \cr} \right) = \left( \matrix{ n + 1 \cr 3 \cr} \right) - \left( \matrix{n \cr 3 \cr} \right) = \left( \matrix{n \cr 2 \cr} \right) = {{n\left( {n - 1} \right)} \over 2} \cr & \sum\limits_{k = 4}^7 {k\left( {k - 1} \right)} = \sum\nolimits_{k = 4}^{7 + 1} {k\left( {k - 1} \right)} = 2\left( {\left( \matrix{ 7 + 1 \cr 3 \cr} \right) - \left( \matrix{ 4 \cr 3 \cr} \right)} \right) = 104 \cr & \sum\limits_{k = a}^b {k\left( {k - 1} \right)} = 2\left( {\left( \matrix{ b + 1 \cr 3 \cr} \right) - \left( \matrix{ a \cr 3 \cr} \right)} \right) \cr} $$

G Cab
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