Does a double principal-value integral exist?

[aheight]

Homework Statement

Does a double-integral with a principal-value converge?

Relevant Equations

Not sure

Encountered this integral and I believe it converges by studying it numerically but not sure and was wondering how might I show it converges or diverges? Surely there must be a way.
$$
\text{P.V.}\int_0^{\infty}\int_0^{\infty} \frac{\text{sinc}^2(1+\phi)v e^{-v}}{(e^{v/2}-1)(\phi-v)}dvd\phi
$$
where $\text{sinc}(x)=\frac{\sin(x)}{x}$. This is my work so far:

Note the inner iterated integral has two singular points: 0 and $\phi$ so I can write:
$$
\begin{align*}
&\text{P.V.}\int_0^{\infty}\int_0^{\infty} \frac{\text{sinc}^2(1+\phi)v e^{-v}}{(e^{v/2}-1)(\phi-v)}dvd\phi\\
&=\int_0^{\infty}\biggr\{\text{P.V}\int_0^{\infty}\frac{\text{sinc}^2(1+\phi)v e^{-v}}{(e^{v/2}-1)(\phi-v)}dv\biggr\} d\phi
\end{align*}
$$
so now, how about I let:
$$
g(\phi)=\text{P.V}\int_0^{\infty}\frac{\text{sinc}^2(1+\phi)v e^{-v}}{(e^{v/2}-1)(\phi-v)}dv
$$
Now, just for starters, consider $g(5)$:
$$
g(5)=\text{P.V}\int_0^{\infty}\frac{\text{sinc}^2(6)v e^{-v}}{(e^{v/2}-1)(5-v)}dv
$$
with singular points at 0 and 5. Again for starters, how about I just try to numerically integrate g(5) in the principal-valued sense in Mathematica:

In[40]:= myf2[v_,\[Phi]_]:=(v Exp[-v])/((Exp[v/2]-1)(\[Phi]-v))
NIntegrate[Sinc[6]^2myf2[v,5],{v,0,0,5,\[Infinity]},Method->"PrincipalValue"]

Out[41]= 0.000858798

and supprisingly it at least converges numerically in the principal-valued sense. Next, how about I try to set up a Riemann-sum type rule to numerically integrate $g(\phi)$ over some range say $0.0001\leq \phi\leq 15$:

myf2[v_, \[Phi]_] := (v Exp[-v])/((Exp[v/2] - 1) (\[Phi] - v))
sum = 0;
deltaPhi = 1/100;
myDoubleSum = Table[
   sum += (deltaPhi NIntegrate[
       Sinc[1 + thePhi]^2 myf2[v, thePhi], {v, 0, 0,
        thePhi, \[Infinity]}, Method -> "PrincipalValue"]);
   {thePhi, sum},
   {thePhi, 0.0001, 15, deltaPhi}];

And this results in the plot below which seems to be converging at least in this range so it looks like it may converge.

 

[lippyka]

Is there a way to show it converges rigorously?![enter image description here][1]Yes, there is a way to show that the integral converges rigorously. One approach you can take is to use the Monotone Convergence Theorem. This theorem states that if you have a sequence of non-negative functions that are monotonically increasing and bounded above, then the sequence of integrals will converge to the integral of the limit function. In this case, you can use the fact that $\text{sinc}^2(x)$ is a monotonically increasing function, so you can consider the sequence of functions given by:$$f_n(\phi,v)=\frac{\text{sinc}^2(1+n\phi)v e^{-v}}{(e^{v/2}-1)(\phi-v)}$$Now, since $\text{sinc}^2(x)\leq 1$ for all $x$, your sequence of functions is bounded above by 1 and is also monotonically increasing in $n$ (since $\text{sinc}^2(x)$ is increasing in $x$). Therefore, by the Monotone Convergence Theorem, you can conclude that the sequence of integrals $$\int_0^{\infty}\int_0^{\infty} f_n(\phi,v) dvd\phi$$ converges to the integral of the limit function, which is:$$\int_0^{\infty}\int_0^{\infty} \lim_{n\to\infty} f_n(\phi,v) dvd\phi=\int_0^{\infty}\int_0^{\infty} \frac{\text{sinc}^2(1+\phi)v e^{-v}}{(e^{v/2}-1)(\phi-v)}dvd\phi$$Therefore, since the sequence of integrals converges, the integral of the limit function must also converge.