Integral of a sum under condition

Question

I'm struggling with the following integral, would anyone have an idea on how to approach it? I solved it for N = 2 but the generalization to any positive integer is more difficult.

$$\int_{(\omega_1, \cdots, \omega_N) \in [0,1]^N \\ \sum_{i=1}^N \omega_i = 1} \sum_{i=1}^N \min{ (\omega_i\cdot E, L_i)}$$

with:

-$E \geqslant 0$

-$\forall i, L_i >0$

-$N > 1$, strictly positive integer

Solution for N = 2:

I found the following for N = 2:

	$E < L_2$	$E > L_2$
$E< L_1$	$E$	$\frac{E}{2} + L_2 - \frac{L_2^2}{2E}$
$E > L_1$	$\frac{E}{2} + L_1 - \frac{L_1^2}{2E}$	$L_1 + L_2 - \frac{L_1^2 + L_2^2}{2E}$

after having computed $\int_0^1 (\min{(\alpha E, L_1)} + \min{((1 -\alpha) E, L_2)} )\,d\alpha $

What is a bit frustating is to not have a nice formula, even for N=2 (this may exist but I haven't found it).

I have a master's degree in applied mathematics but I'm not that young anymore and a bit rusty so I would really appreciate the help of a fresh mind. (But I accept contributions from older persons ;) ).

Kind regards

Answer 1

This is really a comment, not an answer, but my MathJax isn't good enough to type it in a comment box. For $i=1,2$, let $$ f_i=\begin{cases} \frac E2,&E\leq L_i\\ L_i-\frac{L_i^2}{2E},&E>L_i \end{cases}$$

Then when $N=2$, your answer equals $$\sum_{i=1}^2f_i$$

I would try to generalize this approach. See if you can evaluate the integral for a single summand.

Answer 2

First, a rewriting of your formula to show the power of using my $M_i.$

Letting $M_i=\min(L_i/E,1)$ your table becomes:

	$M_2=1$	$M_2<1$
$M_1=1$	$E$	$E\left(\frac{1}{2} + M_2 - \frac{M_2^2}{2}\right)$
$M_1<1$	$E\left(\frac{1}{2} + M_1 - \frac{M_1^2}{2}\right)$	$E\left(M_1 + M_2 - \frac{M_1^2 + M_2^2}{2}\right)$

So in all cases you get one formula which works for all $M_i,$ with no cases:

$$E\left(M_1 + M_2 - \frac{M_1^2 + M_2^2}{2}\right)$$

---

Now, let us compute:

$$E\int_{(\omega_j)\in[0,1]^N\,\sum \omega_1=1} \min(\omega_i,M_i)$$

This can be rewritten:

$$\int_{t\in[0,M_i]}t\int_{(x_j)\in[0,1-t]^{N-1}\,\sum x_j =1-T}1\\+\int_{t\in [M_i,1]} M_i \int_{(x_j)\in[0,1-t]^{N-1}\,\sum x_j =1-t}1$$

Now $$\int_{(x_j)\in[0,1-t]^{N-1}\,\sum x_j =1-t}1=\frac{(1-t)^{N-2}}{(N-2)!}$$

So we get:

$$\frac{E}{(N-2)!}\left(\int_0^{M_i}t(1-t)^{N-2}\,dt+M_i\int_ {M_i}^1 (1-t)^{N-2}\right)\,dt$$

Writing $$t(1-t)^{N-2}=(1-t)^{N-2}-(1-t)^{N-1}$$

You get: $$\frac{E}{(N-2)!}\left(\left(\frac1N(1-t)^N-\frac1{N-1}(1-t)^{N-1}\right)\bigg \vert_0^{M_i} -\frac{M_{i}}{N-1}(1-t)^{N-1}\bigg\vert_{M_i}^1\right)$$

That’s a bit messy, but you get:

$$\frac{E}{(N-2)!}\left(\frac{(1-M_i)^N}{N}-\frac{(1-M_i)^{N-1}}{N-1}+\frac{1}{N(N-1)} +M_i(1-M_i)^{N-1}\right)\\ =\frac{E(1-M_i)^{N-1}}{N!}\left((N-1)^2M_i-1\right)+\frac{E}{N!}$$

So your integral is:

$$\frac{E}{(N-1)!}+\frac{E}{N!}\sum_i (1-M_i)^{N-1}((N-1)^2M_i-1)$$

When $N=2$ this gives:

$$E+\frac E2\sum_i(-1+2M_i-M_i^2)=E\left(M_1+M_2-\frac{M_1^2+M_2^2}{2}\right)$$

When $N=3$ you get:

$$\frac E2+\frac{E}{6}\sum_i (1-M_i)^2(4M_i-1)\\=\frac E6\sum_i\left(4M_i^3-9M_i^2+6M_i\right)$$

—-

You can write $$(N-1)^2M_i-1=N(N-2)-(N-1)^2(1-M_i).$$ That lets you write everything as polynomials of $K_i=1-M_i.$

$$\frac E{(N-1)!}+\frac E{N!}\sum_i \left(N(N-2)K_i^{N-1}-(N-1)^2 K_i^N\right)$$