Items tagged with residue


Greetings to all.

With the following matter I am betting on there being a simple mistake on my part due to fatigue owing to a challenging session of intense computing. The following link at Math.Stackexchange.Com points to a computation involving complex residues. Consult the link for additional details.

I usually verify my computations with Maple, I did the same this time. Thereby I happened on a curious phenomenon which I have documented below. Please study the session data provided, I believe it speaks for itself.

user@host:~$ math
Mathematica 10.0 for Linux x86 (64-bit)
Copyright 1988-2014 Wolfram Research, Inc.

In[1]:= Residue[z^2/(z^4 + 2*z^2 + 2)^2, {z, 2^(1/4)*Exp[3*Pi*I/8]}]

            1/8      1/4
        (-1)    ((-1)    + Sqrt[2])
Out[1]= ----------------------------
            1/4      1/4           3
        16 2    ((-1)    - Sqrt[2])

In[2]:= N[%]

Out[2]= 0.117223 - 0.0083308 I

user@host:~$ maple
    |\^/|     Maple 18 (X86 64 LINUX)
._|\|   |/|_. Copyright (c) Maplesoft, a division of Waterloo Maple Inc. 2014
 \  MAPLE  /  All rights reserved. Maple is a trademark of
 <____ ____>  Waterloo Maple Inc.
      |       Type ? for help.
> residue(z^2/(z^4 + 2*z^2 + 2)^2, z=2^(1/4)*exp(3*Pi*I/8));

> quit
memory used=0.9MB, alloc=8.3MB, time=0.07

I am looking forward to learning what the correct syntax is to get the residue in this case and I hope I can assist other users who might have run into the same problem. I will cancel the question should it turn out to be trivial and of little potential use to the community.

Best regards,

Marko Riedel

Post Scriptum. Being a programmer myself I would be curious to learn more about the algorithms that are deployed here and how and why they did not succeed.

Mathematica 10.3.0 was announced yesterday. This is the 6th release of Mathematica 10 during 16 months. I wonder its  MathematicaFunctionData and   FindFormula . At first sight, the former is an analog of FunctionAdvisor of Maple, but the latter isn't any analog. Also compare the outputs of




>`assuming`([residue(binomial(n, k), n = -j)], [integer, j > 0]);

                residue(binomial(n, k), n = -j)
Let us wait for Maple 2016.


hi  for example to calculate the following

residue((Psi(-z)+Eulergamma)^2*h(z), z = 2)



but it possible to write 

as( Psi(2)+Eulergamma(z))*h(2)+(D(h))(2)

so that 

and Psi(z)+Eulergamma== harmonicNumber(z-1)

the result must be


it is possible that Maple gives explit form of the values function avoid to calculate automatic.




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,


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));

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,...

Dear friends,

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...

Dear friends,

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



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...

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