Resiude Question

  • #1
I have to integrate [itex]\int_{\Gamma} \frac{\log{z+i}}{1+z^2}[/itex] around a large semi-cricular contour.

i found a sinle pole with residue [itex]\frac{\log{2i}}{2i}[/itex] at [itex]z=i[/itex] and a double pole with residue [itex]-\frac{1}{2i}-\frac{\log{2i}}{2i}[/itex] at [itex]z=-i[/itex]. this seems fair enough but i can't decide where to put my contour so as to enclose them both. the reside theorem tells me they will integrate to [itex]-\pi[/itex] if i can get them both inside the contour which is probably a good aim consdiering the next part of the question is to show

[itex]\int_0^{\infty} \frac{\log{x^2+1}}{x^2+1}dx=\pi \log{2}[/itex]

(i) where do i put the contour
(ii) how can i evaluate the next part given that my complex integral is in terms of z's and the one i want is in terms of x^2's so it isn't as simple as taking the real part after i use Jordan's lemma on teh semi circle is it?


Answers and Replies

  • #2
Can you double check that wrote down the problem down correctly, both the first and second parts? Things just seem a little off, maybe not though.

There are a few things.
1) Why is the point z=-i a pole of order 2 and not 1?
2) The countour you will need is [itex]\Gamma = \gamma_R + [-R,-1/R] +\gamma_{1/R} + [r,R][/itex], where [itex]\gamma_R = Re^{it},\,\gamma_{1/R} = -e^{-it}/R[/itex] and [itex]R>0,\,t\in[0,\pi][/itex]. You'll let [itex]R\to\infty[/itex], and note that [itex]1/R\to0[/itex]. Look at the second contour in the below figure.

You need such a contour because you need [itex]\log z[/itex] to be an analytic branch of the logarithm, for example the branch that deletes 0 and the negative imaginary axis.
3) The way to do such an integral is:
\lim_{R\to\infty} \int_\Gamma f(z) \,dz &= \lim_{R\to\infty} \left( \int_{\gamma_R} f(z)\,dz + \int_{-R}^{-1/R} f(x)\,dx + \int_{\gamma_{1/R}} f(z) \,dz + \int_{1/R}^R f(x)\,dx \right) \\
2\pi i \sum_{z_k\in \Gamma} \text{Res}\, (f;z_k) &= \int_{-\infty}^0 f(x)\,dx + \int_0^\infty f(x)\,dx + \lim_{R\to\infty} \left( \int_{\gamma_R} f(z)\,dz + \int_{\gamma_{1/R}} f(z) \,dz \right) \\
The integrals left over on the right are usually shown to tend to zero as [itex]R\to\infty[/itex] or are evaluated otherwise.
4) You don't need all the poles of f(z) to be inside [itex]\Gamma[/itex]. You just worry about the ones that are as [itex]R\to\infty[/itex].
5) Also, if f(x) is even then
\int_{-\infty}^0 f(x)\,dx + \int_0^\infty f(x) \,dx = \int_{-\infty}^\infty f(x) \,dx = 2\int_0^\infty f(x) \,dx


  • Contour_750.gif
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  • #3
it is meant to be log(z+i) and log(x^2+i)
  • #4
you say we need log to be holomorphic so we deleted zer but our contour still contains the negative real axis?

with such a contour, i would only have the simple pole's contribution

i said we had a double pole at z=-i because initially when i went for a simple pole and started evaluating the residues, i got something divided by 0 (which usually means you've missed an order of your pole doesn't it?) so i assumed that log wouldn't be holomorphic at z=-i either and that made it a double pole and the residue calculation then worked fine

can you comment on these three points please
thanks for your help
  • #5
Yo have to avoid the beanch point singularity of the Log (which is not a regular "pole"). The only way to do that is to split Log(x^2+1):

Log(x^2 + 1) = Log[(x+i)(x-i)] = Log(x+i) + Log(x-i)

Then you split up the integral in two parts. For each part you use a different contour. The part containing the Log(x+i) has the branch point singularity at z = -i, so you put the semi-circle above the real axis and for the integral containing the Log(x-i), you put the semi circle below the real axis.

Now, you can also evaluate only one of the two integrals and take twice the real part, because the integral of Log(x-i)/(x^2 + 1) will be the complex conjugate of the integral of Log(x+i)/(x^2 + 1).

This is then a bit similar to evaluating the integral of sin(x)/x using contour integration. You can just replace sin(x) by exp(i x), take the contour with the semi circle in the upper half plane and then take the real part of the final result. But in more complicated cases, simply taking the real part may not work. The general method is to write
sin(x) as [exp(ix) - exp(-ix)]/(2 i), split up the integral in two parts and then take different contours for the two parts.