Edited by Valentin Fadeev, Thursday, 27 Mar 2014, 10:11
As a follow up thought I realized just how much easier it would have been to calculate the residue by definition, i.e. expanding the integrand in the Laurent series to get the coefficient . Let where is small:
Of course, care is needed when choosing the value of the root. It depends on the value of the argument set on the upper bound of the cut. Since I chose it to be 0, the correct value of the root is .
Now
Therefore, near the integrand has the following expansion:
where is the regular part of the expansion which is of no interest in this problem.
Edited by Valentin Fadeev, Thursday, 21 Apr 2011, 23:58
I found this example in a textbook dated 1937 which I use as supplementary material for M828. It gave me some hard time but finally sorted out many fine tricks of contour integration. Some inspiration was provided by the discussion of the Pochhammer's extension of the Eulerian integral of the first kind by Whittaker and Watson.
Evaluate the following integral:
Consider the integral in the complex plane:
where contour C is constructed as follows. Make a cut along the segment of the real axis. Let on the upper edge of the cut. Integrate along the upper edge of the cut in the positive direction. Follow around the point along a small semi-circle in the clockwise direction. The argument of the second factor in the numerator will decrease by . Then proceed along the real axis till and further round the circle in the counter-clockwise direction.
This circle will enclose both branch point of the integrand and , however since the exponents add up to unity the function will return to its initial value:
This is the reason why we only need to make the cut along the segment and not along the entire real positive semi-axis. Then integrate from in the negative direction, then along the second small semi-circle around the point where the argument of the second factor will again decrease by . Finally integrate along the lower edge of the cut and along a small circle around the origin where the argument of the first factor will decrease by .
As the result of this construction the integral is split into the following parts:
Two integrals around the small circles add up to an integral over a circle:
Similarly, the integral around a small circle around the origin vanishes:
Integrals along the segment cancel out:
The integral over the large circle also tends to 0 as R increases. This can be shown using the Jordan's lemma, or by direct calculation:
Finally the only two terms that survive allow us to express the contour integral in terms of the integral along the segment of the real axis:
The contour encloses the only singularity of the integrand which is the pole of the third order at . Hence, by the residue theorem:
The residue can be calculated using the standard formula:
Calculation of the derivative can be facilitated by taking logarithm first:
Edited by Valentin Fadeev, Sunday, 23 Jan 2011, 21:49
Had to do some revision of vector calculus/analysis before embarking on M828.
One point which I was not really missing, but did not quite get to grips with was the double vector product. I remembered the formula:
,
but nevertheless had difficulties applying it in excercises.
The reason whas that that the proof I saw used the expression of vector product in coordinates and comparison of both sides of the equation. However, I was aware of another, purely "vector" argument with no reference to any coordinate system.
Eventually I was able to reproduce only part of it, consulting one old textbook for some special trick. So here's how it goes.
For is perpendicular to the plane of and , must lie in this plane, therefore:
Dot-multiply both parts by :
Since , left-hand side is 0, so:
Now define vector lying in the plane of and , perpendicular to and directed so that , and form the left-hand oriented system. This guarantees that the angle between and , .
Dot-multiply both parts by :
However,
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and
Hence
can be calculated in a similar manner, however, it is easier achieved using equation .
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