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Suppose that $a_1<1$, $a_1+a_2+a_3>1.$ For $x,y,z>0,$

(1) define a fucntion $$H(x,y,z)=\frac{x^{\frac{1}{2}}\int_0^{\infty}\frac{1}{t^{a_1}~ (1+t)^{a_2+1}~ (1+t+z)^{a_3}}\exp\big\{-\frac{x}{1+t}-\frac{ y}{1+t+z}\big\}dt }{\int_0^{\infty}\frac{1}{t^{a_1}~ (1+t)^{a_2}~(1+t+z)^{a_3}}\exp\big\{-\frac{x}{1+t}-\frac{ y}{1+t+z}\big\}dt}.$$ Then $H(x,y,z)$ is uniformly bounded over $x,y$, i.e. there is a constant C, such that $H(x,y)\le C.$

(2) Furthermore, define $$L(y,z)=\frac{y^{\frac{1}{2}}\int_0^{\infty}\frac{1}{t^{a_1}~ (1+t)^{a_2}~ (1+t+z)^{a_3+1}}\exp\big\{-\frac{ y}{1+t+z}\big\}dt }{\int_0^{\infty}\frac{1}{t^{a_1}~ (1+t)^{a_2}~(1+t+z)^{a_3}}\exp\big\{-\frac{y}{1+t+z}\big\}dt}.$$

Then $L(x,y)$ is also uniformly bounded over $x,y.$

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  • $\begingroup$ What is the question? $\endgroup$
    – Fedor Petrov
    Commented Jan 15, 2019 at 20:02
  • $\begingroup$ How to prove uniformly bounded?? Need help. $\endgroup$
    – Xiaopai Song
    Commented Jan 15, 2019 at 20:52
  • $\begingroup$ There seems to be a typo. In the current version of the question, the numerator of your fraction is just $x^{1/2}$ times the denominator $\endgroup$
    – Yemon Choi
    Commented Jan 15, 2019 at 21:00
  • $\begingroup$ Thank you. Actually, I think it is right. The power of (1+t) is different. $\endgroup$
    – Xiaopai Song
    Commented Jan 15, 2019 at 21:08

1 Answer 1

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We may assume that $x>100(1+|a_2|+|a_3|)^2$, for other $x$ simply $C=10(1+|a_2|+|a_3|)$ works perfectly.

Partition the integral in the numerator onto two parts: $I_1$ over $(0,\sqrt{x}]$ and $I_2$ over $[\sqrt{x},\infty)$. The second part does not exceed the denominator, since $\sqrt{x}(1+t)^{-1}<1$ for all $t\geqslant \sqrt{x}$. It suffices to estimate $I_1\leqslant c I_2$ for some fixed $c>0$ (depending only on $a_1,a_2,a_3$). For this it suffices to prove the pointwise estimate $f(t)\leqslant 2^{a_1} f(2t)$ for all $t\in (0,\sqrt{x})$ where $f$ is the integrated function. Indeed, integrating this over $(0,\sqrt{x})$ we get $\int_0^{\sqrt{x}}f(t)dt\leqslant 2^{a_1}\int_0^{\sqrt{x}}f(2t)dt=2^{a_1-1}\int_0^{2\sqrt{x}} f(t)dt$, thus $I_1(1-2^{a_1-1})\leqslant 2^{a_1-1}I_2$ as desired. We have $$ \frac{f(t)}{2^{a_1}f(2t)}=\left(\frac{1+2t}{1+t}\right)^{a_2+1}\cdot \left(\frac{1+2t+z}{1+t+z}\right)^{a_3}\cdot \exp\left(-\frac{x}{1+2t}\left(\frac{1+2t}{1+t}-1\right)- \frac{z}{1+2t+z}\left(\frac{1+2t+z}{1+t+z}-1\right) \right). $$ We estimate $\frac{1+2t+z}{1+t+z}\leqslant \frac{1+2t}{1+t}$ and use the estimate $1+s\leqslant e^s$ for $s:=\frac{1+2t}{1+t}-1$. Also estimate $-\frac{z}{1+2t+z}\left(\frac{1+2t+z}{1+t+z}-1\right)\leqslant 0$ in the exponent. We get $$ \frac{f(t)}{2^{a_1}f(2t)}\leqslant \exp\left(\left(1+|a_2|+|a_3|-\frac{x}{1+2t}\right)s\right)\leqslant 1, $$ since $\frac{x}{1+2t}\geqslant \frac{x}{1+2\sqrt{x}}>\frac13\sqrt{x}>1+|a_2|+|a_3|$ due to our assumptions.

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  • $\begingroup$ Your answer is amazing!!!. Applying your method, I try to prove the following similar question. But I failed. Could you help me? $\endgroup$
    – Xiaopai Song
    Commented Jan 16, 2019 at 18:22
  • $\begingroup$ I have updated the question. Question (2) is similar. I am not sure your method still works. I fail to prove it by your method. $\endgroup$
    – Xiaopai Song
    Commented Jan 16, 2019 at 18:30
  • $\begingroup$ @xiaopai833 I think, everything works: we break the integral at the point $t_0$ for which $\sqrt{y}/(1+t_0+z)=100A$, where $A=|a_2|+|a_3|+1$, (if such a point does not exist, the ratio is clearly bounded.) On $(0,t_0]$ we estimate the ratio $\frac{f(t)}{2^{a_1}f(2t)}$ using the estimate $(1+2t)/(1+t)\leqslant 1+t\leqslant e^t$. And $tA\leqslant y(\frac1{1+t+z}-\frac1{1+2t+z})=\frac{ty}{(1+t+z)(1+2t+z)}$. $\endgroup$
    – Fedor Petrov
    Commented Jan 16, 2019 at 19:19
  • $\begingroup$ Yes. Everything still works. $\endgroup$
    – Xiaopai Song
    Commented Jan 17, 2019 at 13:23
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    $\begingroup$ I am afraid that it is no longer true if say $a_1=1/2,a_2=2,a_3=0$ (I say now about $L$), $b_1=1$, $b_2=100000000$. Then the integral in the denominator over $[0,1]$ looks to be much greater than all three integrals over $[0,1],[1,1+z]$ and $[1+z,\infty)$ in certain regime for $y$ and $1+z$. Within a bounded factor, you may significantly simplify these integrals (say, on $[0,1]$ you replace $1+t$ to 1, $1+t+z$ to $1+z$, $b_1y/(1+t+z)$ to $y/(2+z)$; on $[1,1+z]$ in the denominator you replace $1+t$ to $t$, $1+t+z$ to $z$, $e^{-b_2y/(1+t+z)}$ to $e^{-50y/(1+z)}$ etc.) $\endgroup$
    – Fedor Petrov
    Commented Jan 17, 2019 at 19:35

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