# Does a double principal-value integral exist?

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

Out= 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##:
Mathematica:
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. Last edited:
• Delta2