- #1

aheight

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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:

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##:

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

$$

\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
```

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: