Integration by residual theorem

In summary: But that's not my remark. My remark is that if you think about it, the principal value is not unique. I do know that it is a ##\pi## value because of the ##\pi## under the ##\int_{-R}^R##. So I will still try to find an antiderivative, even though the ##\pi## is not a part of the limit.I would have to look at my old notes to see how to do it, but I think you can get rid of the ##e^{ix}## and get a simpler problem before integrating by parts. (But I could be wrong.)I have no idea how to do the contour integral, which is
  • #1
AlonsoMcLaren
90
2

Homework Statement



Integrate [tex] \int_{-\infty}^{\infty} dx \ x^{-3}(sin x -x )[/tex]


Homework Equations



Residual theorem, Jordan's lemma

The Attempt at a Solution



[tex] \int_{-\infty}^{\infty} dx \ x^{-3}(sin x -x )

= I am \ (PV\int_{-\infty}^{\infty} dx \ x^{-3}(e^{ix} -ix ))[/tex]

In the contour, if we let R goes to infinity and ρ goes to zero, then we have

[tex]\oint_{Total}^{ } dz \ z^{-3}(e^{iz} -iz) = \int_{C_{1}}^{ } dz \ z^{-3}(e^{iz} -iz) + \int_{C_{2}}^{ } dz \ z^{-3}(e^{iz} -iz) + PV \int_{-\infty}^{\infty} dx \ x^{-3}(e^{ix} -ix) [/tex]

Because the only singularity is 0, not inside the contour. By residual theorem (or Cauchy's theorem),
[tex]\oint_{Total}^{ } dz \ z^{-3}(e^{iz} -iz) = 0[/tex]

By Jordan's lemma, as R goes to infinity,

[tex] \int_{C_{2}}^{ } dz \ z^{-3}(e^{iz} -iz) =0[/tex]

For the C1 contour

[tex] \int_{C_{1}}^{ } dz \ z^{-3}(e^{iz} -iz)= \int_{\pi}^{0 } d(\rho e^{i\theta }) \frac{e^{i\rho e^{i\theta }} -i\rho e^{i\theta }}{(\rho e^{i\theta })^{3}}= i \int_{\pi}^{0 }\frac{e^{i\rho e^{i\theta }}-i\rho e^{i\theta}}{\rho ^{2}e^{2i\theta }} d \theta[/tex]

But as ρ goes to 0, [tex]\frac{e^{i\rho e^{i\theta }}-i\rho e^{i\theta}}{\rho ^{2}e^{2i\theta }} [/tex]

goes to infinity. So I am stuck.

The final answer, according to WolframAlpha, is -π/2
 

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  • #2
AlonsoMcLaren;4530276 But as ρ goes to 0 said:
\frac{e^{i\rho e^{i\theta }}-i\rho e^{i\theta}}{\rho ^{2}e^{2i\theta }} [/tex]

goes to infinity.

No.

Your mission, should you choose to accept it, is to show:

[tex]\lim_{\rho\to 0} \int_{\pi}^{0}\frac{e^{i\rho e^{it}}-i\rho e^{it}}{\rho^2 e^{2 i t}} dt =\pi/2[/tex]

and you can't just say it is cus' I told you so because I could be wrong. I've not been able to do so myself yet. Ain't that the fun part of math?

Edit: Ok, I think I have it. Kinda' rickety and not sure about the legitimacy of applying L'Hopital's rule to integral functions but it's something. You try to get it and let me know what you get.
 
Last edited:
  • #3
jackmell said:
No.

Your mission, should you choose to accept it, is to show:

[tex]\lim_{\rho\to 0} \int_{\pi}^{0}\frac{e^{i\rho e^{it}}-i\rho e^{it}}{\rho^2 e^{2 i t}} dt =\pi/2[/tex]

and you can't just say it is cus' I told you so because I could be wrong. I've not been able to do so myself yet. Ain't that the fun part of math?

Edit: Ok, I think I have it. Kinda' rickety and not sure about the legitimacy of applying L'Hopital's rule to integral functions but it's something. You try to get it and let me know what you get.

A PF member was kind enough to help me with a part of this analysis in the Calculus sub-forum and according to his analysis, I am justified in applying L'Hopital's rule under the integral sign in this particular case:

https://www.physicsforums.com/showthread.php?p=4531822&posted=1#post4531822

Fun problem. Thanks voko!
 
  • #4
It is clear that the original integral converges since ##x^{-3}## will govern. And I suppose you are taking a course in complex variables and are supposed to use contour integrals.

But in a contrarian spirit, allow me to suggest that you can find an antiderivative for this integrand. Integrate by parts several times.

And in an inquiring spirit, could you enlighten me as to what PV means. I know the Cauchy Integral formula but have not seen the notation PV.

Re applying L'Hospital's theorem, that theorem is just expressing some facts about the Taylor's polynomial. No reason why you can't expand your function under an integral sign (we do it all the time).
 
  • #5
brmath said:
But in a contrarian spirit, allow me to suggest that you can find an antiderivative for this integrand. Integrate by parts several times.

How would you do this? I assume you intend to integrate the x-3 but when you run into a log you are going to be in trouble.

And in an inquiring spirit, could you enlighten me as to what PV means. I know the Cauchy Integral formula but have not seen the notation PV.

It means principal value, and means instead of taking the true double sided limit you take
[tex] \lim_{R\to \infty} \int_{-R}^{R} [/tex]
which is what you actually get when you do contour integration (meaning you can do contour integration to calculate the principal value of something where the true integral doesn't exist). If the integral converges then it converges to the same value as the principal value of course.
 
  • #6
jackmell said:
No.

Your mission, should you choose to accept it, is to show:

[tex]\lim_{\rho\to 0} \int_{\pi}^{0}\frac{e^{i\rho e^{it}}-i\rho e^{it}}{\rho^2 e^{2 i t}} dt =\pi/2[/tex]

But [tex]\lim_{\rho\to 0} \int_{\pi}^{0}\frac{e^{i\rho e^{it}}-i\rho e^{it}}{\rho^2 e^{2 i t}} dt =\pi/2[/tex] is indeed infinity

If you do taylor expansion:

[tex]e^{i\rho e^{it}} = 1+ i\rho e^{it}-\frac{1}{2}\rho^{2} e^{2it}+...[/tex]

[tex]e^{i\rho e^{it}}-i\rho e^{it} = 1-\frac{1}{2}\rho^{2} e^{2it}+...[/tex]

[tex]\frac{e^{i\rho e^{it}}-i\rho e^{it}}{\rho^2 e^{2 i t}} = \rho^{-2}e^{-2it}-1/2+..[/tex]

It blows up as ρ goes to zero!
 
  • #7
Office_Shredder said:
How would you do this? I assume you intend to integrate the x-3 but when you run into a log you are going to be in trouble.



It means principal value, and means instead of taking the true double sided limit you take
[tex] \lim_{R\to \infty} \int_{-R}^{R} [/tex]
which is what you actually get when you do contour integration (meaning you can do contour integration to calculate the principal value of something where the true integral doesn't exist). If the integral converges then it converges to the same value as the principal value of course.

Sorry about my contrarian spirit, which both saw that the ##x^{-3}## would govern, and viewed it as ##x^3## for integration purposes. This is too contrarian even for me.

I've used lim##\int_{-R}^R## in contour integration, but didn't know it was called the Principal Value.
 
  • #8
AlonsoMcLaren said:
But [tex]\lim_{\rho\to 0} \int_{\pi}^{0}\frac{e^{i\rho e^{it}}-i\rho e^{it}}{\rho^2 e^{2 i t}} dt =\pi/2[/tex] is indeed infinity

If you do taylor expansion:

[tex]e^{i\rho e^{it}} = 1+ i\rho e^{it}-\frac{1}{2}\rho^{2} e^{2it}+...[/tex]

[tex]e^{i\rho e^{it}}-i\rho e^{it} = 1-\frac{1}{2}\rho^{2} e^{2it}+...[/tex]

[tex]\frac{e^{i\rho e^{it}}-i\rho e^{it}}{\rho^2 e^{2 i t}} = \rho^{-2}e^{-2it}-1/2+..[/tex]

It blows up as ρ goes to zero!

Outstanding Alonso. I see you've completed your mission and taught me a very simple way to show the limit:

[tex]\begin{align*}
\lim_{\rho\to 0}\int_{\pi}^0 \frac{e^{i\rho e^{it}}- i\rho e^{it}}{\rho^2 e^{it}}dt&=\lim_{\rho\to 0}\int_{\pi}^0 \left(-\frac{1}{2}+\frac{e^{-2 i t}}{\rho ^2}-\frac{1}{6} i e^{i t} \rho +\frac{1}{24} e^{2 i t} \rho ^2+\frac{1}{120} i e^{3 i t} \rho ^3+\cdots\right)dt \\
&=\pi/2+\lim_{\rho\to 0} \frac{1}{\rho^2}\int_{\pi}^0 e^{-2 i t}dt+0 \\
&=\pi/2+\lim_{\rho\to 0} \frac{1}{\rho^2}(0)+0 \\
&=\pi/2
\end{align*}
[/tex]
 
Last edited:

1. What is the residual theorem?

The residual theorem is a mathematical concept that is used in integration to evaluate the value of an integral by finding the residues of a function at its singular points. It is a useful tool for solving integrals that cannot be evaluated using traditional methods.

2. How is the residual theorem used in integration?

The residual theorem is used in integration by converting a complex integral into a contour integral, which can then be evaluated using the residues of the function. The residues are found by taking the limit of the function as it approaches a singular point.

3. What are singular points in integration?

Singular points in integration are points on a complex plane where a function is not continuous. These points can be poles or branch points, and they can affect the behavior of a function around them. The residual theorem is used to evaluate integrals around these points.

4. Can the residual theorem be used for any integral?

No, the residual theorem is only applicable for integrals that have singular points. If a function does not have any singular points, then the residual theorem cannot be used to evaluate the integral.

5. What are some applications of the residual theorem?

The residual theorem has many applications in mathematics, physics, and engineering. It is used to solve integrals in complex analysis, to calculate the coefficients of Laurent series, and to evaluate contour integrals in physics and engineering problems.

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