As part of a larger proof, I am currently trying to prove the following equation: $$\int_0^r \int_0^{r-t_1} \dots \int_0^{r-\sum_{i=1}^{k-1} t_i} t_1^{m_1 - 1} t_2^{m_2 - 1} \dots t_k^{m_k - 1} dt_k \dots dt_1 = \frac{r^n}{n!} \prod_{i=1}^{k} (m_i -1)!$$

I know that $n = \sum_{i=1}^k m_i$ and moreover that all $m_i$ are natural numbers.

I've already confirmed with Mathematica that this equation holds for $k \in \{1,\dots,5\}$. I however have some difficulties with proving this equation for any $k > 0$. Any hints would be greatly appreciated!

Here's what I've done so far:

In order to get started, I tried to show it for $k = 1$: $$\int_0^r t_1^{m_1 - 1} dt_1 = \left[\frac{1}{m_1} \cdot t_1^{m_1}\right]_{t_1=0}^{t_1=r} = \frac{r^{m_1}}{m_1} = \frac{r^{m_1}}{m_1!} \cdot \frac{m_1!}{m_1} = \frac{r^{m_1}}{m_1!} \cdot (m_1 - 1)!$$

That was pretty straightforward.

But already for $k=2$ I'm getting stuck:

$$\int_0^r \int_0^{r-t_1} t_1^{m_1 - 1}t_2^{m_2 - 1} dt_2dt_1 = \int_0^r t_1^{m_1 - 1}\int_0^{r-t_1} t_2^{m_2 - 1} dt_2dt_1 = \int_0^r t_1^{m_1 - 1}\left[\frac{1}{m_2} \cdot t_2^{m_2}\right]_{t_2=0}^{t_2=r-t_1}dt_1 = \int_0^r t_1^{m_1 - 1}\cdot\frac{1}{m_2} \cdot (r-t_1)^{m_2}dt_1$$

Now if I would like to integrate over $t_1$, I should probably use the binomial theorem, which says: $$(x+y)^n = \sum_{k=0}^n {n \choose k}x^{n-k}y^k = \sum_{k=0}^n {n \choose k}x^{k}y^{n-k}$$ Applied to my case, this results in: $$\begin{align*} &\int_0^r t_1^{m_1 - 1}\cdot\frac{1}{m_2} \cdot \sum_{j=0}^{m_2} {{m_2} \choose j} \cdot r^{m_2-j}\cdot (-1)^j \cdot t_1^jdt_1\\ &= \frac{1}{m_2} \cdot \int_0^r \sum_{j=0}^{m_2} (-1)^j \cdot\frac{m_2!}{j!(m_2-j)!} \cdot r^{m_2-j}\cdot t_1^{m_1 + j - 1}dt_1\\ &=\frac{m_2!}{m_2} \cdot \sum_{j=0}^{m_2} \left((-1)^j \cdot\frac{1}{j!(m_2-j)!} \cdot r^{m_2-j} \int_0^r t_1^{m_1 + j - 1}dt_1\right)\\ &=(m_2-1)! \cdot \sum_{j=0}^{m_2} \left((-1)^j \cdot\frac{1}{j!(m_2-j)!} \cdot r^{m_2-j} \left[\frac{1}{m_1 + j} \cdot t_1^{m_1 + j}\right]_{t_1=0}^{t_1=r}\right)\\ &=(m_2-1)! \cdot \sum_{j=0}^{m_2} (-1)^j \cdot\frac{1}{j!(m_2-j)!} \cdot r^{m_2-j} \cdot \frac{1}{m_1 + j} \cdot r^{m_1 + j}\\ &=(m_2-1)! \cdot \sum_{j=0}^{m_2} (-1)^j \cdot\frac{1}{j!(m_2-j)!} \cdot r^{m_2+m_1} \cdot \frac{1}{m_1 + j}\\ &=r^{m_1 + m_2}(m_2-1)! \cdot \sum_{j=0}^{m_2} (-1)^j \cdot\frac{1}{j!(m_2-j)!}\cdot\frac{1}{(m_1 + j)} \end{align*}$$ Now the only thing left to show is that the sum is equal to $\frac{(m_1-1)!}{(m_1+m_2)!}$. I tried the following: $$\begin{align*} &\sum_{j=0}^{m_2} (-1)^j \cdot\frac{1}{j!(m_2-j)!}\cdot\frac{1}{(m_1 + j)}\\ &= \sum_{j=0}^{m_2} (-1)^j \cdot\frac{1}{j!(m_2-j)!}\cdot\frac{m_1 \cdots (m_1+j-1)\cdot(m_1+j+1)\cdots(m_1+m_2)}{m_1 \cdots (m_1 + m_2)} \cdot \frac{(m_1-1)!}{(m_1-1)!}\\ &= \frac{(m_1-1)!}{(m_1+m_2)!} \cdot \sum_{j=0}^{m_2} (-1)^j \cdot\frac{1}{j!(m_2-j)!} \cdot \left(\prod_{i=1, i \neq j}^{m_2} (m_1 + i)\right) \end{align*}$$ So now I only need to show that the remaining sum is equal to one, but I'm not able to do so. Again, playing around with some examples ($m_2 \in \{1,2,3\}$) everything works out. I've tried to reframe it as $\frac{(m_1-1)!}{(m_1+m_2)!m_2!} \cdot \sum_{j=0}^{m_2} (-1)^j \cdot{m_2 \choose j}\cdot \left(\prod_{i=1, i \neq j}^{m_2} (m_1 + i)\right)$, but not with much success.

I'm not quite sure whether I'm on the right track here or whether there's a different approach to this whole thing that looks more promising than what I'm trying to do. I feel like I won't be able to prove the overall thing for $k>0$ if I'm not even able to show it for the special case $k=2$.

I've been working on this for a couple of days now and I don't seem to make progress. If you have any ideas what I could try, that would be great. I'd be very grateful for both help with the $k=2$ proof and for ideas about how to tackle the general $k>0$ proof. Thanks in advance!


Here is a result for $k=2$ (EDIT) and for $k=3$ which shows the general way to proceed.

You are looking at imcomplete beta functions. For $k=2$, I follow your result (calling that $I_2$):

$$ I_2 = \int_0^r t_1^{m_1 - 1}\cdot\frac{1}{m_2} \cdot (r-t_1)^{m_2}{\rm d}t_1 $$

Substituting $t_1 = r z$ gives

$$ I_2 = \frac{r^{m_1+m_2}}{m_2} \int_0^1 z^{m_1 - 1}\cdot (1-z)^{m_2} {\rm d} z $$

Now use the definition for the incomplete beta function $B(\cdot,\cdot)$ and its special values for natural numbers:

$$ I_2 = \frac{r^{m_1+m_2}}{m_2} B(m_1, m_2+1) = \frac{r^{m_1+m_2}}{m_2} \frac{(m_1 - 1)! m_2 !}{(m_1+m_2)!} = r^{m_1+m_2} \frac{(m_1 - 1)! (m_2 -1) !}{(m_1+m_2)!} $$

which is the desired result for $k=2$.

One step further, for $k=3$. The $r$'s have been substituted away as above.

$$ I_3 = \frac{r^n}{m_3} \int_0^1 z_1^{m_1 - 1} \int_0^{1-z_1} z_2^{m_2 - 1}\cdot (1-z_1-z_2)^{m_3} {\rm d}z_2{\rm d}z_1 $$

Now for the inner integral, substitute $z_2 = s_2 (1-z_1)$ which gives $$ I_3 = \frac{r^n}{m_3} \int_0^1 z_1^{m_1 - 1} (1-z_1)^{m_2+m_3} \int_0^{1} s_2^{m_2 - 1}\cdot (1-s_2)^{m_3} {\rm d}s_2{\rm d}z_1 $$

This is a formulation where the inner integral has been made independent on the outer one - and this will work out for larger $k$ as well!

Making this explicit, we get $$ I_3 = \frac{r^n}{m_3} \Big[\int_0^1 z_1^{m_1 - 1} (1-z_1)^{m_2+m_3} {\rm d}z_1 \Big] \cdot \Big[\int_0^{1} s_2^{m_2 - 1}\cdot (1-s_2)^{m_3} {\rm d}s_2\Big] $$

Now using the already known results for the Beta function above, $$ I_3 = \frac{r^n}{m_3} \frac{(m_1 - 1)! (m_2 + m_3) !}{(m_1+m_2+m_3)!} \cdot \frac{(m_2 - 1)! m_3 !}{(m_2+m_3)!} = {r^n} \frac{(m_1 - 1)! (m_2 - 1)! (m_3 -1)!}{(m_1+m_2+m_3)!} $$

as desired.

From here, you can go on to larger $k$.

  • $\begingroup$ Thanks a lot, the variable change and the Beta function really do the trick! I'll try to formulate the proof for general $k$ but it really looks like it's not as difficult as I thought. $\endgroup$
    – user432709
    Apr 7 '17 at 9:42

So building on the solution of @Andreas for $k=2$ and $k=3$, I was able to prove the whole thing for arbitrary $k>0$:

First of all, one can observe that for any natural number $j > 0$ the following holds: $$\int_0^{r-\sum_{i=1}^{j-1}t_i} t_j^{a-1} \cdot \left(r- \sum_{i=1}^j t_i\right)^b dt_j = B(a,b+1) \cdot \left(r- \sum_{i=1}^{j-1} t_i\right)^{a+b}$$ We can define $t_j = \left(r- \sum_{i=1}^{j-1} t_i\right)\cdot z$ which gives $dt_j = \left(r- \sum_{i=1}^{j-1} t_i\right)\cdot dz$. Making a variable change in the left part of the equation results in $$\int_0^{1} \left(r- \sum_{i=1}^{j-1} t_i\right)^{a-1} \cdot z^{a-1} \cdot \left(r- \sum_{i=1}^{j-1} t_i - \left(r- \sum_{i=1}^{j-1} t_i\right)\cdot z\right)^b \cdot\left(r- \sum_{i=1}^{j-1} t_i\right) dz$$ $$= \left(r- \sum_{i=1}^{j-1} t_i\right)^{a-1+b+1} \int_0^{1} z^{a-1} (1-z)^b dz = B(a,b+1) \cdot \left(r- \sum_{i=1}^{j-1} t_i\right)^{a+b}$$

Now we look at the overall problem: $$I_k = \int_0^r \int_0^{r-t_1} \dots \int_0^{r-\sum_{i=1}^{k-1} t_i} t_1^{m_1 - 1} t_2^{m_2 - 1} \dots t_k^{m_k - 1} dt_k \dots dt_1$$ This can be rewritten as $$I_k = \int_0^r t_1^{m_1 - 1} \int_0^{r-t_1} t_2^{m_2 - 1} \dots \int_0^{r-\sum_{i=1}^{k-1} t_i} t_k^{m_k - 1} dt_k \dots dt_1$$

Using the observation made above, we can solve the innermost integral by setting $j=k, a= m_k, b=0$ which gives us $B(m_k,1) \cdot \left(r- \sum_{i=1}^{k-1} t_i\right)^{m_k}$

We thus get: $$I_k = B(m_k,1) \cdot \int_0^r t_1^{m_1 - 1} \dots \int_0^{r-\sum_{i=1}^{k-2} t_i} t_{k-1}^{m_{k-1} - 1} \cdot \left(r- \sum_{i=1}^{k-1} t_i\right)^{m_k} dt_k \dots dt_1$$

Again, we apply our observation to the innermost integral ($j=k-1,a=m_{k-1},b={m_k}$, resulting in $B(m_{k-1},m_k+1) \cdot \left(r- \sum_{i=1}^{k-2} t_i\right)^{m_{k-1}+m_k}$

Recursively applying this step finally results in: $$I_k = B(m_k,1)\cdot B(m_{k-1},m_k + 1) \cdot B(m_{k-2},m_{k-1}+m_k + 1)\cdot \dots \cdot B(m_1, m_2+\dots+m_k+1)\cdot r^{m_1+\dots+m_k}$$ As all $m_i$ are natural numbers, we can use that $B(x,y)=\frac{(x-1)!\,(y-1)!}{(x+y-1)!}$: $$I_k = r^{m_1+\dots+m_k} \cdot \frac{(m_k-1)!\,0!}{m_k!} \cdot \frac{(m_{k-1}-1)!\,m_k!}{(m_{k-1}+m_k)!} \cdot \frac{(m_{k-2}-1)!\,(m_{k-1}+m_k)!}{(m_{k-2}+m_{k-1}+m_k)!} \cdot \dots \cdot \frac{(m_1-1)!\,(m_2+\dots+m_k)!}{(m_1+m_2+\dots+m_k)!}$$ This reduces to: $$I_k = r^{m_1+\dots+m_k} \cdot (m_k -1)! \cdot \dots \cdot (m_1 -1)! \cdot \frac{1}{(m_1+\dots+m_k)!}$$ With $m_1+\dots+m_k = n$, we can rewrite this as: $$I_k = \frac{r^n}{n!} \prod_{i=1}^{k} (m_i -1)!$$

  • $\begingroup$ Nice! That's the general way to proceed. $\endgroup$
    – Andreas
    Apr 7 '17 at 13:38

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