# Does a double principal-value integral exist?

• aheight
In summary: This shows that the given integral is convergent. In summary, by using the Monotone Convergence Theorem and the fact that ##\text{sinc}^2(x)## is monotonically increasing and bounded above, we can show that the given integral is convergent.
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:
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##:
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
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.

## 1. What is a double principal-value integral?

A double principal-value integral is an integral that involves two singular points within the integration limits. These singular points are typically poles, where the function being integrated becomes infinite.

## 2. How is a double principal-value integral different from a regular integral?

A regular integral can be evaluated using standard integration techniques, while a double principal-value integral requires a more specialized approach. This is because the singular points within the integration limits make the integral diverge, so it cannot be evaluated using traditional methods.

## 3. When does a double principal-value integral exist?

A double principal-value integral exists when the singular points within the integration limits cancel each other out, resulting in a finite value for the integral. This can happen when the function being integrated has symmetrical singularities or when the singularities are of opposite signs.

## 4. How is a double principal-value integral calculated?

A double principal-value integral is typically calculated using Cauchy's principal value theorem, which involves taking the limit of the integral as the singular points approach each other. This can be a complex and time-consuming process, so it is often done using numerical methods.

## 5. What are some real-world applications of double principal-value integrals?

Double principal-value integrals are commonly used in physics and engineering to solve problems involving electric and magnetic fields, as well as in quantum mechanics to calculate the energy levels of particles. They are also used in signal processing and image reconstruction techniques.

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