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Let $t_1,\ldots,t_m>0$, and $m\ge 4$ be an even integer. Consider the function: $$ f(a,b;\mathbf{t})=\int_0^{t_1}\ldots\int_0^{t_m} |x_1-x_m|^a |x_2-x_1|^b |x_3-x_2|^a |x_4-x_3|^b \ldots |x_{m-1}-x_{m-2}|^a |x_{m}-x_{m-1}|^b d\mathbf{x}. $$ (It is finite when $a,b>-1$ and $a+b>2(1/m-1)$. See 1.)

The goal is to study the asymptotics of $f(a,b;\mathbf{t})$ as $(a,b)\rightarrow(0,-1)$.

Heuristically, by first setting $a=0$, and then using the identity (for $b>-1$): $$ \int_0^{t_1}\int_0^{t_2}|x_1-x_2|^{b} dx_1dx_2=\frac{1}{(b+1)(b+2)} \left( t_1^{b+2}+ t_2^{b+2}-|t_1-t_2|^{b+2} \right), $$ one expects as $(a,b)\rightarrow(0,-1)$ that $$ f(a,b;\mathbf{t})\sim (b+1)^{-m/2} \prod_{i=2,4,...m} \left( t_i+ t_{i-1}-|t_{i}-t_{i-1}| \right). $$ But I had a hard time justifying this (if it is correct) rigorously. The Dominated Convergence seems not directly applicable because as $b\rightarrow -1$ the integrand is not bounded by an integrable function.

Could any one offer some idea? Many thanks.

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    $\begingroup$ The answer is correct. To get it, instead of thinking how large everything is, think of how large/small part of the whole it is. By the way, $a+b-|a-b|$ is also known as $2\min(a,b)$. $\endgroup$
    – fedja
    Jul 4, 2014 at 2:14
  • $\begingroup$ Hi @fedja. Thank you for your attention. Could you give me more specific hints? I would really appreciate it. $\endgroup$
    – Uchiha
    Jul 4, 2014 at 6:36
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    $\begingroup$ You shouldn't have any trouble with the lower bound because the estimate $|u|^{-a}\ge 1-$ is uniform on compacts. The same could be said about the upper bound if not for a small neighborhood of $0$. So, split $|u|^{-a}=1+h$ where $h$ is small in $L^1$ and start looking at multiple integrals with at least one $h$ in them. Increase all intervals to $[0,T]$ and decrease all distances to $T$-periodic distances. Then you get a full convolution and can compute everything, but the bound for each $h$-term will still be negligible compared to the main term coming from all $1$'s. Should I say more? $\endgroup$
    – fedja
    Jul 4, 2014 at 16:14
  • $\begingroup$ Sorry @fedja. I just could not see how a full convolution is obtained due to the circular structure. Can you explain further? $\endgroup$
    – Uchiha
    Jul 5, 2014 at 0:47
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    $\begingroup$ You mean that you have your chain cyclic rather than linear? Yes, this is a minor nuisance I should have paid more attention to, but if you just consider any two available $a$-th powers in $L^2$ instead of $L^1$ and recall that, when convolving one $L^2$ function with plenty of $L^1$ ones, you still get a good control of the $L^2$-norm of the result, you can split the chain by Cauchy-Schwarz. Sorry for overlooking this small detail (without which it is, indeed, not clear why $m\ge 4$ matters) :) $\endgroup$
    – fedja
    Jul 5, 2014 at 4:34

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