Assume that $a_1<1$, $a_3<1,$ $a_1+a_2+a_5>1$, $a_3+a_4+a_5>1,$ $a_1+a_2+a_3+a_4+a_5>2.$ For $x,y>0,$ define a fucntion $$H(u,v)=\frac{u^{\frac{1}{2}}\int_0^{\infty}\int_0^{\infty}\frac{1}{x^{a_1}~ (1+x)^{a_2+1}~ y^{a_3}~(1+y)^{a_4}~ (1+x+y)^{a_5}}\exp\big\{-\frac{u}{1+x}-\frac{v}{1+x+y}\big\}dx dy}{\int_0^{\infty}\int_0^{\infty}\frac{1}{x^{a_1}~ (1+x)^{a_2}~ y^{a_3}~(1+y)^{a_4}~(1+x+y)^{a_5}}\exp\big\{-\frac{2u}{1+x}-\frac{2v}{1+x+y}\big\}dx dy}.$$ Then $H(u,v)$ is uniformly bounded over positive constant $u,v$, i.e. there is a constant C, such that $H(u,v)\le C.$

How to prove it? Fedor Petrov gave the answer [Uniformly Bounded (updating) for a similar case. I need your help.

Maybe, the following result is useful:

If $a_1<1$ and $a_1+a_2>1,$ then $$f(u)\equiv\int_0^{\infty}\frac{1}{x^{a_1}~ (1+x)^{a_2}}\exp\big\{-\frac{u}{1+x}\big\}dt\approx \min\{1,u^{1-a_1-a_2}\}.$$

  • $\begingroup$ @Fedor Petrov are you interested in this question? $\endgroup$ – Xiaopai Song Mar 10 at 16:18
  • $\begingroup$ It looks bit technical, but I think that partitioning of integral onto several pieces onto which the integrated function may be simplified should either answer your question or at least reduce it to something more pretty. Did you try it? $\endgroup$ – Fedor Petrov Mar 12 at 12:37
  • $\begingroup$ @Fedor Petrov You mean partitioning of integral onto several pieces , for example, [0,1],[1,N],[N,\infty) over x and y? Do you have suggestions for the choice of partitioning? Let me try. $\endgroup$ – Xiaopai Song Mar 13 at 13:36

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