Suppose that $a, b$ and $c$ are constant.

Is there the necessary and sufficient conditions of $a ,b, c$ for the following integration is integrable? i.e. $$\int_0^\infty \int_0^\infty \int_0^\infty \frac{1}{(1+x)^a (1+y)^b (1+x+t)^c (1+y+t)^c} t^{-\frac{1}{2}}e^{-\frac{1}{t}} \, dx \, dy \, dt < \infty.$$


closed as off-topic by Alexandre Eremenko, abx, Ben McKay, Neil Hoffman, Pace Nielsen Jan 22 at 21:36

This question appears to be off-topic. The users who voted to close gave these specific reasons:

  • "MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics." – Ben McKay, Neil Hoffman, Pace Nielsen
  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Alexandre Eremenko, abx
If this question can be reworded to fit the rules in the help center, please edit the question.


Let us assume that $a,b,c$ are real numbers. The integral in question equals \begin{equation} I=I_1+I_2, \end{equation} where \begin{equation} I_1:=\int_0^2 dt\; t^{-1/2}e^{-1/t} J_a(t)J_b(t),\quad I_2:=\int_2^\infty dt\; t^{-1/2}e^{-1/t} J_a(t)J_b(t), \end{equation} \begin{equation} J_a(t):=\int_0^\infty\frac{dx}{(1+x)^a(1+x+t)^c}. \end{equation} Clearly, if $I<\infty$, then $J_a(t)<\infty$ for almost all $t>0$, whence $a+c>1$. Similarly, it is necessary that $b+c>1$. So, without loss of generality \begin{equation} a+c>1,\quad b+c>1. \tag{1} \end{equation}

So, for $t\in(0,2]$, \begin{equation} J_a(t)\asymp\int_0^\infty\frac{dx}{(1+x)^a(1+x)^c}\asymp1, \end{equation} whence one always has \begin{equation} I_1\asymp1<\infty, \end{equation} given the necessary condition (1).

For $t>2$, \begin{align} J_a(t)&\asymp\int_0^1\frac{dx}{t^c}+\int_1^t\frac{dx}{x^a t^c}+\int_t^\infty\frac{dx}{x^a x^c} \\ &\asymp t^{-c}+t^{-c}\int_1^t\frac{dx}{x^a}+t^{-c+1-a}, \end{align} whence \begin{equation} J_a(t)\asymp \begin{cases} t^{(1-a)_+-c}&\text{ if } a\ne1,\\ t^{-c}\ln t&\text{ if } a=1. \end{cases} \end{equation} Since $e^{-1/t}\asymp1$ for $t>2$, it follows that \begin{equation} I_2<\infty\iff (1-a)_+-c+(1-b)_+-c-1/2<-1 \iff c>\tfrac14+(1-a)_+ +(1-b)_+. \end{equation} In particular, the latter condition implies $c>(1-a)_+\ge1-a$ and similarly $c>1-b$, so that the necessary condition (1) holds.

Thus, for $I<\infty$ it is necessary and sufficient that \begin{equation} c>\tfrac14+(1-a)_+ +(1-b)_+. \end{equation}

  • $\begingroup$ Your answer for Problem (1) is right. I have updated the question. Based on your ideal, I can't handle question (2). Could you help me to fix the question (2)? $\endgroup$ – Xiaopai Song Jan 21 at 15:19
  • 1
    $\begingroup$ @xiaopai833 : If you want another question answered, in addition to your original post, you should ask it in a separate post. $\endgroup$ – Iosif Pinelis Jan 21 at 18:46
  • $\begingroup$ I am so sorry. I'm a beginner. As you said, I have posted another question. If you have time, please help me. [mathoverflow.net/q/321407/134602] $\endgroup$ – Xiaopai Song Jan 21 at 18:58

Not the answer you're looking for? Browse other questions tagged or ask your own question.