Complex Integral: Solve for pi.a coth(2.pi.a) - 1/2

In summary: Thanks for your input!In summary, Belgium 12 is having trouble with a complex integral and is looking for help. He tried different contours and methods, but without success. He suggested trying the integral with just the answer, but that didn't work either. Finally, he suggests trying sin²(a.lnz)/(x-1)²=-1/4(z.exp 2ia -2 +z.exp -2ia)/(x-1)², which will give him the result z.exp(2ia+b) -2.z.exp(b) -z.exp(-2ia+b).
  • #1
Belgium 12
43
0
Hi,

I have a problem with the following complex integral.

Integral from 0 to+infinty sin²(a.lnz)/(x - 1)² = pi.a coth(2.pi.a) - 1/2 a>0

I tried different contours and methods,but without result.

Can you help me to find out the complex contour integration.

Thanks
 
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  • #2
Do you mean

[tex]
\int_0^\infty{}\frac{\sin^2{(a \log z)}}{(z-1)^2}
[/tex]
with z in the denominator instead of x. What contour did you try?
 
  • #3
Welcome to PF!

Hi Belgium 12! Welcome to PF! :smile:

(do you mean (z - 1)² ?)

Hint: sin²θ = [1 - cos2θ]/2; and cos(2a ln(z)) = … ? :smile:
 
  • #4
Yes that's the integral.

I tried a contour in upper half plane with an identitation at 1.

Then key-hole contour with the branch 0<phi<2.pi.

The a rectangular contour after the subtitution z= exp u

tanks
 
  • #5
tiny-tim said:
Hi Belgium 12! Welcome to PF! :smile:

(do you mean (z - 1)² ?)

Hint: sin²θ = [1 - cos2θ]/2; and cos(2a ln(z)) = … ? :smile:
This would be

[tex]
\frac{1}{2}\left[e^{2ia\log z}+e^{-2ia\log z}\right]=\frac{1}{2}\left[z^{2ia}+z^{-2ia}\right]
[/tex]

How does this help?
 
  • #6
Hi,

I try it out and come later with a possible answer
 
  • #7
Hi,

I have no result with cos(2alnz)=1/2(z exp2ia + z exp -2ia)

you have then sin²=(1-cos 2 phi)/2

thus 1/2{1- 1/2(zexp 2ia+zexp -2ia)}/ (x-1)²

We already have a second order pole on the real axis (x-1)²/2

I used the key-hole contour.

thanks
 
  • #8
hi,
the result of the integral pi.a coth(2pia)-1/2 is correct.I saw it in a book "integral tables"

The integral with just the answer comes from a book in french "Recueil de problemes

sur la theorie des fonctions analytiques" by Evgrafov MIR publishers Moscow.
 
  • #9
Hi,
tiny tim

Maybe I should try sin²(a.lnz)/(x-1)²=(exp 2ialnz -2 +exp2ialnz).-1/4/(x-1)²

ans with the term -2/(x-1) partial integration in the PV sence.

int from 1+e to infinty +int from 0 to 1-e.
Then the two other integrals 1)zexp2ia/(x-1)² and zexp -2ia/(x-1)² and the adding.

what's your opinion?

thanks
 
  • #10
Hi Belgium 12! :smile:

Yes, that's what I was thinking of originally … it gives you something like:
[tex]\frac{\left[z^{2ia}+z^{-2ia}\,-\,2\right]}{4 (z\,-\,1)^2}\,.[/tex]​
Unfortunately, as Pere Callahan asks, does that actually get us anywhere?

I'm very rusty on contour integrals.

I suspect it has something to do with a quadrant integral, going down the positive real axis, up the positive imaginary axis, and then round in a huge 90º arc. But that still needs some way of (literally!) getting round the pole at z = 1. :confused:

Sorry … I'm stuck. :redface:
 
  • #11
Hi,
Tiny Tim
Callehan
I found this in a bib,in a french book .Integral with double pole on the real axis.
It's a bit difficult for me for writing the math.expressions.I don't have latex or an other program.

Sorry for that
 

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  • #12
Hi Belgium,

I have to admit your file is too long for me to read right now ... and the math expression don't speed up the reading. Why don't you try learning latex (it's nothing about the keyboard, just some markup language, similar to HTML) and post your specific question about the method (if there are any)?
 
  • #13
Hi,
Pere callahan

About your answer ysterday evening about the integral.I would say its not a specifc

method.Ist just a complex contour integration,but you don't see it many books on

complex integration.

The method is just they insert a convergence factor.

sin²(a.lnz)/(x-1)²=-1/4(z.exp 2ia -2 +z.exp -2ia)/(x-1)² they insert the factor let us say

z.expb

or -1/4(z.exp(2ia+b) -2.z.exp(b) -z.exp(-2ia+b))/(x-1)².

then integrate the three individual integrals take the sum en let b->0

that was it

Belgium 12
 
  • #14
an then employ the formula

{int from o to 1-r xexp.a/(x-1)² + int from 1+r to inf. xexp.a/(x-1)² - 2/r}=-pi.a.cotg(pi.a)

to the three terms a=2ia and a=-2ia
 
  • #15
Hi Belgium 12! :smile:

Is this supposed to be part of thread https://www.physicsforums.com/showthread.php?t=227979&goto=newpost ? :smile:
 
  • #16
yes that's is.

Belgium 12
 
  • #17
I've merged the threads.
 

What is a complex integral?

A complex integral is a mathematical concept that involves finding the area under a curve in the complex plane. It is similar to a regular integral in real calculus, but it deals with functions that have complex values.

How do you solve for pi in a complex integral?

In order to solve for pi in a complex integral, you need to manipulate the equation to isolate the pi variable on one side. Then, you can use algebraic techniques to solve for pi, such as factoring or substitution. In some cases, you may need to use more advanced techniques, such as contour integration.

What does the coth function represent in a complex integral?

The coth function stands for the hyperbolic cotangent, which is defined as the ratio of the hyperbolic cosine to the hyperbolic sine. In a complex integral, the coth function is often used to simplify the integration process, as it is a combination of two common trigonometric functions.

Why is the constant 1/2 included in the equation for a complex integral?

The constant 1/2 is included in the equation to account for the difference between the complex integral and the regular integral. In order to accurately solve for pi in a complex integral, this constant must be taken into consideration.

How can complex integrals be used in scientific research?

Complex integrals have many applications in scientific research, particularly in fields such as physics, engineering, and mathematics. They are often used to solve problems involving complex functions and to model real-world phenomena. For example, they can be used to calculate the electric potential of a charged particle or to analyze the behavior of signals in a communication system.

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