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.
$$\int\_0^\infty \int\_0^\infty \int\_0^\infty \frac{1}{(1+x)^a (1+y)^b (1+x+t)^c (1+y+t)^c} t^{-\frac{1}{2}}e^{-\frac{1}{t}} \, dx \, dy \, dt < \infty.$$



Necessary and Sufficient conditions for integrable function
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}


