help with an irregular integral


by mmzaj
Tags: integral, irregular
mmzaj
mmzaj is offline
#1
Dec9-12, 01:11 PM
P: 99
I am looking for help with doing the following integral :
[tex]\frac{1}{2\pi i}\int_{1}^{\infty}\ln\left(\frac{1-e^{-2\pi i x}}{1-e^{2\pi i x}} \right )\frac{dx}{x\left(\ln x+z\right)}\;\;\;\;z\in \mathbb{C}[/tex]
i tried to transform it into a complex integral along a 'keyhole' contour, with a branch cut along the +ive real line [itex]\left[1,\infty\right)[/itex]. but then [itex]\;\ln x\;[/itex] would be transformed into [itex]\;\ln x+2\pi i\;[/itex] when doing the integral along [itex]\left(\infty,1\right]\;[/itex] which doesn't add up nicely to the portion along [itex]\left[1,\infty\right)[/itex]. any insights are appreciated.
Phys.Org News Partner Science news on Phys.org
SensaBubble: It's a bubble, but not as we know it (w/ video)
The hemihelix: Scientists discover a new shape using rubber bands (w/ video)
Microbes provide insights into evolution of human language
haruspex
haruspex is offline
#2
Dec9-12, 11:35 PM
Homework
Sci Advisor
HW Helper
Thanks ∞
P: 9,215
Doesn't this simplify fairly easily?
Quote Quote by mmzaj View Post
[tex]\frac{1-e^{-2\pi i x}}{1-e^{2\pi i x}}[/tex]
mmzaj
mmzaj is offline
#3
Dec10-12, 03:06 AM
P: 99
you would think !!! but no, it doesn't ....

mmzaj
mmzaj is offline
#4
Dec10-12, 10:09 AM
P: 99

help with an irregular integral


The integral above is equivalent to :

[tex] \int_{1}^{\infty}\left(\frac{1}{2}-x+\left \lfloor x \right \rfloor \right )\left(\frac{1}{x\left(\ln x+z\right)}\right)dx [/tex]
And
[tex]\int_{1}^{\infty}\sum_{n=1}^{\infty}\frac{\sin(2 \pi n x)}{n\pi}\left(\frac{1}{x\left(\ln x+z\right)}\right)dx [/tex]
haruspex
haruspex is offline
#5
Dec10-12, 03:56 PM
Homework
Sci Advisor
HW Helper
Thanks ∞
P: 9,215
[itex]\frac{1-e^{-2\pi i x}}{1-e^{2\pi i x}} = \frac{e^{-2\pi i x}(e^{2\pi i x}-1)}{1-e^{2\pi i x}} = -e^{-2\pi i x}[/itex]
I end up with [itex]\int_0^∞\frac{1-2e^u}{2(u+z)}du[/itex]
which surely doesn't converge?
mmzaj
mmzaj is offline
#6
Dec12-12, 01:20 PM
P: 99
Quote Quote by haruspex View Post
[itex]\frac{1-e^{-2\pi i x}}{1-e^{2\pi i x}} = \frac{e^{-2\pi i x}(e^{2\pi i x}-1)}{1-e^{2\pi i x}} = -e^{-2\pi i x}[/itex]
I end up with [itex]\int_0^∞\frac{1-2e^u}{2(u+z)}du[/itex]
which surely doesn't converge?
you missed the fact that the inverse of the complex exponential - the complex [itex]\log[/itex] function- is multivalued. namely :
[tex] \frac{1-e^{-2\pi i x}}{1-e^{2\pi i x}}=-e^{-2\pi i x}=e^{-2\pi i \left(x-1/2\right)}=e^{-2\pi i \left(\left \{ x \right \}-1/2\right)}[/tex]
Where [itex] \left \{ x \right \}[/itex] is the fractional part of x. Thus:
[tex] \frac{1}{2\pi i}\ln\left(\frac{1-e^{-2\pi i x}}{1-e^{2\pi i x}}\right)=\frac{1}{2}-\left \{ x \right \}[/tex]
Another way to think of it is to take the Taylor expansion of the [itex] \log[/itex]:
[tex] \frac{1}{2\pi i}\left(\ln\left(1-e^{-2\pi i x}\right)-\ln(1-e^{2\pi i x})\right)=\frac{1}{2\pi i}\sum_{k=1}^{\infty}\frac{e^{2\pi i k x}-e^{-2\pi i kx}}{k}=\sum_{n=1}^{\infty}\frac{\sin(2\pi k x)}{k\pi}[/tex]
Which in turn is the Fourier expansion of [itex]\frac{1}{2}-\left \{ x \right \} [/itex]
Using these facts, we can prove that the integral in question is equal to the limit:
[tex] e^{-z}\text{Ei}(z)+\lim_{N\rightarrow \infty}\sum_{n=1}^{N}\left(n+\frac{1}{2} \right )\ln\left(\frac{\ln(n+1)+z}{\ln(n)+z} \right )-e^{-z}\text{Ei}(z+\ln N)[/tex]
Where [itex] \text{Ei}(z)[/itex] is the exponential integral function. But i'm stuck with this cumbersome limit


Register to reply

Related Discussions
Solving an irregular tetrahedron General Math 10
Centroid of an irregular shape Introductory Physics Homework 2
CM of a Irregular Object Introductory Physics Homework 2
Irregular Cone Geometry Precalculus Mathematics Homework 1
LQC into irregular cosmo Beyond the Standard Model 2