Let $$J\_1=\int\_0^1\frac{1}{\sqrt{1-t\_2}}dt\_2,$$
$$J\_2=\int\_0^1 \int\_0^{t\_2}\frac{1}{\sqrt{1-t\_2}}(\frac{1}{\sqrt{1-t\_3}}+\frac{1}{\sqrt{t\_2-t\_3}})dt\_3dt\_2,$$
$J\_3=\int\_0^1 \int\_0^{t\_2}\int\_0^{t\_3}\frac{1}{\sqrt{1-t\_2}}(\frac{1}{\sqrt{1-t\_3}}+\frac{1}{\sqrt{t\_2-t\_3}})(\frac{1}{\sqrt{1-t\_4}}+\frac{1}{\sqrt{t\_2-t\_4}}+\frac{1}{\sqrt{t\_3-t\_4}})dt\_4t\_3t\_2,$
and so on.


Is it true that $$J\_N\leq C^N\quad \text{for some }C\;?$$


Mathematica and Maple both tells me that $J\_1=2$, $J\_2=2+\pi$ and $J\_3=4+\frac{10\pi}{3}$, but they fail to compute $J\_4$. Any help will be appreciated.




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ps: The largest of the $N!$ terms in $J\_N$ is exactly
$$\int\_0^1 \int\_0^{t\_2}\cdots\int\_0^{t\_N}\prod\_{i=1}^N\frac{1}{\sqrt{t\_i-t\_{i+1}}}dt\_{N+1}\cdots dt\_3dt\_2=\frac{\pi^{N/2}}{\Gamma(\frac{N+2}{2})}$$
which is too large, while the smallest term is 
$$\int\_0^1 \int\_0^{t\_2}\cdots\int\_0^{t\_N}\prod\_{i=1}^N\frac{1}{\sqrt{1-t\_{i+1}}}dt\_{N+1}\cdots dt\_3dt\_2=\frac{2^N}{N!}$$



Rate of growth of an explicit integral
The growth rate is exponential. Specifically, I will prove the bound $J\_n \leq \frac{(n+1)^{n} 2^n}{(n+1)!}$, which grows like $(2e)^n$. I can also show that $J\_n \geq 2^n$. I would guess that neither of these are the true rate of growth.


I'll also give an explicit recursion for the $J\_n$, which makes it clear that they are all in $\mathbb{Q}[\pi]$ and should make it easy to compute them recursively.


It is convenient to define $J\_n(b)$ to be $J\_n$ with every upper bound $1$ replaced by $b$. For example, 
$$J\_2(b) = \int\_{t\_2=0}^b d t\_2 \int\_{t\_3=0}^{t\_2} d t\_3 \frac{1}{\sqrt{b-t\_2}} \left( \frac{1}{\sqrt{b-t\_3}} + \frac{1}{\sqrt{t\_2-t\_3}} \right)$$
By homogeneity, $J\_n(b) = J\_n \cdot b^{n/2}$. We set $J\_0(b)=1$.


Here is the recursion in the form which is easiest to prove:
$$J\_n(b) = \sum\_{k\geq 1} \frac{1}{k!} \sum\_{\begin{matrix} n\_1+n\_2 + \cdots + n\_k = n \\ n\_1, n\_2, \ldots, n\_k >0 \end{matrix}} \prod\_{i=1}^k \int\_{c\_i=0}^b \frac{J\_{n\_1-1}(c\_i) d c\_i}{\sqrt{b-c\_i}}$$


**Sketch of proof of recursion:** The left hand side is a sum of $n!$ terms which are naturally indexed by functions $f: \{ 2,3,\ldots, n+1 \} \to \{1,2,\ldots, n \}$ obeying $f(i) \lt i$. The term corresponding to $f$ has integrand $1/\prod \sqrt{t\_{f(i)} - t\_i}$, where $t\_1=b$.


Group together those summands for which $f^{-1}(1)$ has size $k$; let $(u\_1, u\_2, \ldots, u\_k) = f^{-1}(1)$. For any $x \in \{2,3, \ldots, n \}$, there is precisely one $u\_i$ such that $u\_i \in \{ x, f(x), f(f(x)), f^3(x), \ldots \}$ Say that "$x$ passes through $u\_i$". Let $n\_i$ be the number of $x$ which pass through $u\_i$. We group together all terms on the left with the same *unordered* list of multiplicities $(n\_1, n\_2, \ldots, n\_k)$. On the right, group together all terms which differ only by permuting $(n\_1, n\_2, \ldots, n\_k)$. I claim that the corresponding sums on the two sides match up.


An example is probably clearer than a proof here. Take $n=3$ and look at the partition $3=2+1$. On the left hand side, we have terms accounting for $(f(2), f(3), f(4))$ equal to: $(1, 1, 3)$, $(1,1,2)$ and $(1,2,1)$. On the right hand side, we have $2$ copies of 
$$\frac{1}{2!} \int\_{c\_1=0}^b \int\_{c\_2=0}^b \int\_{t=0}^{c\_1} \frac{d c\_1 dc\_2 dt}{\sqrt{b-c\_1} \sqrt{b-c\_2} \sqrt{c\_1-t}}$$
The two copies cancel the $1/2!$. We are integrating over the region $0 \leq t \leq c\_1$, $0 \leq c\_2$. There are three possible orderings for the integrands: $0 \leq c\_2 \leq t \leq c\_1$, $0 \leq t \leq c\_2 \leq c\_1$ and $0 \leq t \leq c\_1 \leq c\_2$, and these correspond to the three functions $f$ above. 


The confusing thing in writing out a general proof is noticing that the $1/k!$ and the number of orderings of the partition $n\_1+n\_2+ \cdots + n\_k$ exactly cancel to eliminate any double counting. Since you tagged this with "feynmann-integrals", I assume you are used to the way these sort of symmetry factors work. $\square$ 


**Simplifying the recursion** We first plug in $J\_m(b) = J\_m \cdot b^{m/2}$ to obtain
$$J\_n = \sum\_{k\geq 1} \frac{1}{k!} \sum\_{\begin{matrix} n\_1+n\_2 + \cdots + n\_k = n \\ n\_1, n\_2, \ldots, n\_k >0 \end{matrix}} \prod\_{i=1}^k J\_{n\_1-1} \int\_{c\_i=0}^1 \frac{ d c\_i \ c\_i^{(n\_i-1)/2}}{\sqrt{b-c\_i}}$$


The integrals are now simple one dimensional integrals and can be evaluated:
$$J\_n = \sum\_{k\geq 1} \frac{1}{k!} \sum\_{\begin{matrix} n\_1+n\_2 + \cdots + n\_k = n \\ n\_1, n\_2, \ldots, n\_k >0 \end{matrix}} \prod\_{i=1}^k \left( J\_{n\_1-1} \cdot \frac{\sqrt{\pi}\ \Gamma((n\_i+1)/2)}{\Gamma((n\_i+2)/2)} \right)$$
Note that $\left( \frac{\sqrt{\pi}\ \Gamma((n\_i+1)/2)}{\Gamma((n\_i+2)/2)} \right)$ is rational if $n\_i$ is odd and is a rational multiple of $\pi$ if $n\_i$ is even, so this shows that all your polynomials are polynomials in $\pi$. (I am feeling nicely superior to Mathematica now. )


**Upper bounds** We have $ \frac{\sqrt{\pi}\ \Gamma((n\_i+1)/2)}{\Gamma((n\_i+2)/2)} \leq 2$. So $J\_n$ is bounded by $K\_n$, where $K\_n$ is defined by the recursion
$$K\_n = \sum\_{k\geq 1} \frac{1}{k!} \sum\_{\begin{matrix} n\_1+n\_2 + \cdots + n\_k = n \\ n\_1, n\_2, \ldots, n\_k >0 \end{matrix}} \prod\_{i=1}^k \left( 2 \cdot K\_{n\_1-1} \right)$$
Define the generating function $g(x) = \sum\_{n=0}^{\infty} K\_n x^n$. Then this shows
$$g(x) = \exp(2 x g(x))$$


We thus see that $g(x) = -W(-2x)/(2x)$ where $W$ is the [Lambert $W$ function](http://en.wikipedia.org/wiki/Lambert_W_function) and so 
$$K\_n = \frac{(n+1)^{n} 2^n}{(n+1)!}$$
as promised.


**Lower bound** We can use a similar trick to get a lower bound. Note that $\frac{\sqrt{\pi} \Gamma((n+1)/2)}{\Gamma((n+2)/2)} \geq \frac{2}{n}$. So $J\_n \geq L\_n$ where $L$ is defined recursively by $L\_0=1$ and
$$L\_n = \sum\_{k\geq 1} \frac{1}{k!} \sum\_{\begin{matrix} n\_1+n\_2 + \cdots + n\_k = n \\ n\_1, n\_2, \ldots, n\_k >0 \end{matrix}} \prod\_{i=1}^k \left( \frac{2}{n} L\_{n\_1-1} \right)$$


Putting $L(x) = \sum\_{n \geq 0} L\_n x^n$, we see that
$$L(x) = \exp\left( 2 \int\_{t=0}^x L(t) dt \right).$$
This integral equation has the unique solution $L(x) = 1/(1-2x)$, so $L\_n = 2^n$, as promised. 


