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Help with an irregular integral

by mmzaj
Tags: integral, irregular
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mmzaj
#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.
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haruspex
#2
Dec9-12, 11:35 PM
Homework
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P: 9,645
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
#3
Dec10-12, 03:06 AM
P: 99
you would think !!! but no, it doesn't ....

mmzaj
#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
#5
Dec10-12, 03:56 PM
Homework
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P: 9,645
[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
#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


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