# Complex integral

1. Apr 10, 2008

### Belgium 12

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

2. Apr 10, 2008

### Pere Callahan

Do you mean

$$\int_0^\infty{}\frac{\sin^2{(a \log z)}}{(z-1)^2}$$
with z in the denominator instead of x. What contour did you try?

3. Apr 10, 2008

### tiny-tim

Welcome to PF!

Hi Belgium 12! Welcome to PF!

(do you mean (z - 1)² ?)

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

4. Apr 10, 2008

### Belgium 12

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. Apr 10, 2008

### Pere Callahan

This would be

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

How does this help?

6. Apr 10, 2008

### Belgium 12

Hi,

I try it out and come later with a possible answer

7. Apr 11, 2008

### Belgium 12

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. Apr 11, 2008

### Belgium 12

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. Apr 12, 2008

### Belgium 12

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.

thanks

10. Apr 12, 2008

### tiny-tim

Hi Belgium 12!

Yes, that's what I was thinking of originally … it gives you something like:
$$\frac{\left[z^{2ia}+z^{-2ia}\,-\,2\right]}{4 (z\,-\,1)^2}\,.$$​
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.

Sorry … I'm stuck.

11. Apr 16, 2008

### Belgium 12

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. Apr 17, 2008

### Pere Callahan

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. Apr 18, 2008

### Belgium 12

Hi,
Pere callahan

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. Apr 18, 2008

### Belgium 12

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. Apr 18, 2008

### tiny-tim

16. Apr 19, 2008

### Belgium 12

yes that's is.

Belgium 12

17. Apr 19, 2008

### HallsofIvy

Staff Emeritus