Bonjour

Comment se calcul le résidue d'une fonction f(x,y) ?

Si par exemple f(x,y)=1/x+1/y+5/(x*y), alors le residue de f en (0,0) est défini comme étant (1/2*Pi)^2*int(f(x,y)) avec intégrale est sur une sphère qui contient l'origine.

Est ce qu'il est égal au coefficient 5 de 1/(x*y) dans le developpement de laurent en deux variables ??

Merci d'avance,

Gérard.

How to calculate the integral of 1/sqrt(4*z^2+4z+3) over the unit circle counterclockwise for each branch of the integrand? Of course, with Maple.

Greetings to all. I will keep this brief and to the point. I would like to point out a certain deficit in the integral transform package. I have recently been calculating some Mellin transforms at this link and the base functions are of the following type.

g := (p, q) -> 1/(x+p)/(x+(p*q-1)/q);

Now to see the deficit here are some Mellin transforms that...

I will post here the output of two commands that has me a bit worried. Of course it is entirely possible that I am overlooking something very simple, which is why I am writing to your site.

> residue(1/(2^s-1), s=2*Pi*I*3/log(2)); 0

Dear friends,

I am reporting with a brief comment concerning the integral int(1/(1+x^a), x=0..infinity) with a>=2 a real number. This was evaluated here.

Now Maple 15 (X86 64 LINUX) will quite happily compute this in its most simple form involving the sine when a is not a positive integer or a rational number. If it is, however, a beta function term results,...

I will present a series of commands for you to ponder that should speak for themselves.

> q := n -> int(1/(1-2^n*exp(I*theta*n))^2*2*I*exp(I*theta), theta=0..2*Pi); 2 Pi / | 2 I exp(theta I) q := n -> | ------------------------ dtheta | n 2 / (1 - 2 exp(theta n I...

I ran into another problem while using Maple to do residue calculus. The following call

residue(1/x^(1/3)/(x^2+2*x*cos(phi)+1), x=-exp(I*phi));

returns zero -- it does not recognize the residue. On the other hand, if I do a subsitution like this

subs(x=-exp(I*phi), 1/x^(1/3)/(x^2+2*x*cos(phi)+1));

followed by

convert(%, trig); simplify(%);

then I get a divide by zero error, which shows that the...

I was working on the integral of cosh(ax)/cosh(x) from zero to infinity with "a" being a rational number p/q where p<q and p-q being odd. I was quite shocked to discover that Maple 15 (X86 64 LINUX)cannot do this integral while Mathematica gets it right every time. I certainly hope this will be rectified soon.

For those who might be interested the calculation of the integral

there is a link

How to find the residue of {e^(az)}/z(z+b)^m at the pole -b of order m

I'm sorry if I'm duplicating the information here. I tried to post a comment but it doesn't seem to be appearing.

When I change = to := it does not work. Maple says

Error, invalid input: MultiSeries:-series expects its 2nd argument, eqn, to be of type {name, name = algebraic}, but received 1.*I

Hello,

I have a huge expression involving psi and phi called A_1 which I paste below. I am trying to integrate it using the residue theorem in the variable phi. I make a substitution in u as below for sin(k*phi) and ask for the singularities of each operand at u. But when I get my final result it still has the variable u in it. If its evaluated the integral in u then it should not return u in the result. Evidently it’s done something...

Hi guys I am trying to evaluate an integral of the form

A_0 := (1/(2*Pi)+Q*cos(k*phi))^2/(1+(1/(2*Pi)+Q*cos(k*phi))^2);

In this case I specify a value for Q=15 as given below and since maple cannot evaluate the integral directly I have written code to evaluate the integral using the residue theorem. So for each singularity I calculate the residue at that point and multiply the sum of the residues by 2*Pi*I as in the code below. However the...

Okay so I can see how to get the poles. But I have some further questions.

1) For beta AND Q both GREATER THAN 0 WHICH poles have positive imaginary part and is there a PROGRAMMATIC way of selecting these ones?

2) How do I calculate the residue at these poles? The approach given by Alec above seems useful and is much...

> I would like to integrate the following expression that I call s2

> s2 := (1/2/Pi+2*Q*u/(1+u^2))^2/((1+beta^2*(1/2/Pi+2*Q*u/(1+u^2))^2)*(1+u^2));

using the residue theorem. To do this I first want to find the roots (poles) of the expression. I do this by solving for the denominator equal to zero as below.

> s3:=(1+beta^2*(1/2/Pi+2*Q*u/(1+u^2))^2)*(1+u^2);

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