Is it possible to get Maple to show me step by step how to solve a complex contour integral?
f := (x,y,z,v) -> (x+I*x*cos(v)+I*y*sin(v))^(-2)
int(f(x,y,z,v),v=0..2*Pi) assuming(x,real,y,real,z,real,v,real)
But I would like to know how Maple solves this step by step. I tried using the tutor...
So further from this, my original problem was to calculate the so called Whittaker contour integral
so we start from
$$f=\int_{0}^{2\pi} \frac{1}{(x+izcos(\vartheta)+iysin(\vartheta))^{2}}d\vartheta$$
This should give $$f=2\pi/r^(3)$$
could someone show me how? Essentially i changed from...
Ok thanks for the replies everyone....
So the way I obtained the integral in the first place was to go from
$$\vartheta \rightarrow \lambda,$$ by using $$\lambda = e^{i\vartheta}$$
So i guess the contour is an integration over the unit circle like Zinq said.
Ok thanks for your comments
I am trying to teach myself complex analysis . There seems to be multiple ways of achieving the same thing and I am unsure on which approach to take, I am also struggling to visualise the problem...Would someone show me step by step how to solve for example...
Pretty much, but where the solution to the Laplace equation is the addition of some function ##g(w)## and ##h(\tilde{w})## where ##\tilde{w}## is the complex conjugate of ##w##...
The more I'm reading about it, it seems as if I need to look at twistor theory, global and local solutions etc...I...
So It is well known that the 2D solution to the Laplace equation can be obtained by changing to complex coordinates ##u=x+iy## and ##v=x-iy##. This can be extended to n dimensions as long as the complex coordinates chosen also solve the Laplace equation. For example in 3D...
Ok thanks for your response. With regard to the first integral in the first post, you actually can show that they are both equivalent,
##\int (x+y)(dx+dy)=\int xdx +\int xdy+ \int ydx +\int ydy##
from the relationship
##dw=dx+dy##
we also get
##dx=dw-dy##, ##dy=dw-dx##
if you substitute...
However you can do the substitution for example in complex analysis where ##u=x+iy## and ##v=x-iy## with the differentials becoming ##du=dx+idy## and ##dv=dx-idy##, Is this substitution and hence the differentials valid because of the imaginary numbers? and hence when you integrate you can...
so
## \int wdw = \frac{1}{2}w^{2} ##
now if w=x+y,
## \int (x+y)(dx+dw)= \int xdx + \int ydx + \int xdy + \int ydy ##
which can be evaluated and gives
##\int(x+y)(dx+dy)=\frac{1}{2}x^{2}+2xy+\frac{1}{2}y^{2}##
but
##\frac{1}{2}x^{2}+2xy+\frac{1}{2}y^{2} \neq \frac{1}{2}w^{2}##
can...
Hi sorry the subscript on the function was to represent a two 2d vector field. Is there a way to obtain an analytical solution if the function g is known?
How do I go about solving a differential equation of the form
\partial_{x}F_{x}(x,y) + \partial_{y}F_{y}(x,y) = g(x,y)
Where g(x,y) is a known function and I wish to solve for F. I thought i could apply the method of characteristics but the characteristic equation is dependent on coefficients...
Basically I am trying to lorentz transform the magnetic field along θ of a bunch particles which have a gaussian distribution to the radial electric field. However the magnetic field in θ is dependent on the longitiudinal distribution.
Now initially i thought we would just use the standard LT...