Discover MakerZone

MATLAB and Simulink resources for Arduino, LEGO, and Raspberry Pi

Learn more

Discover what MATLAB® can do for your career.

Opportunities for recent engineering grads.

Apply Today

Thread Subject:
need help on essential singularity

Subject: need help on essential singularity

From: duc

Date: 26 Mar, 2011 17:54:04

Message: 1 of 2

i need to integrate the following indefinite function
exp((-(x-y)^2)/2*y^2)/sqrt(2*pi*y^2) ws.r.t y

As it has an essential singularity i understand i need to integrate it numerically but i'm not sure which method will work, and if so which one is most effective in dealing with such functions

your help would be much appreciated

duc

Subject: need help on essential singularity

From: Roger Stafford

Date: 26 Mar, 2011 21:48:04

Message: 2 of 2

"duc " <duc_tran8@yahoo.com> wrote in message <iml97s$792$1@fred.mathworks.com>...
> i need to integrate the following indefinite function
> exp((-(x-y)^2)/2*y^2)/sqrt(2*pi*y^2) ws.r.t y
>
> As it has an essential singularity i understand i need to integrate it numerically but i'm not sure which method will work, and if so which one is most effective in dealing with such functions
>
> your help would be much appreciated
>
> duc
- - - - - - - -
  When you write -(x-y)^2)/2*y^2, that puts the y^2 in the numerator. Is that what you mean or do you mean -(x-y)^2)/(2*y^2) with the y^2 in the denominator? I am guessing you actually mean to have y^2 in the denominator.

  With that supposition, let us assume that x > 0 and define a change of variable:

 t = (x-y)/y

which is to say

 y = x/(t+1).

Then we can base the integral at y = 0 and obtain

 int('exp(-(x-y)^2/(2*y^2))/sqrt(2*pi*y^2)','y',0,y) =
 int('1/sqrt(2*pi)*exp(-t^2/2)/(t+1)','t',(x-y)/y,+inf)

  There is a pole singularity in this latter integral as the lower limit, (x-y)/y, approaches -1 but for any (x-y)/y > -1, it is convergent. This shows that the original integral in y based at y = 0 is well-defined for finite y >= 0 but diverges as y approaches plus infinity.

  The above transformation to t not only indicates the nature of the singularity involved but also enables one to express the integral as a function of the single variable (x-y)/y as a lower limit without an explicit x being present in the integrand. Any of the standard quadrature routines can accomplish an evaluation of the above integral in t numerically with the proviso that the upper limit must stop short of infinity but be large enough to provide the necessary accuracy. You cannot of course start at t = -1 as a lower limit.

  With x or y in the negative range, the situation is similar to the above.

  If my guess above in the first paragraph was wrong, your integral will be divergent as y approaches zero and you will need to base it somewhere else.

Roger Stafford

Tags for this Thread

No tags are associated with this thread.

What are tags?

A tag is like a keyword or category label associated with each thread. Tags make it easier for you to find threads of interest.

Anyone can tag a thread. Tags are public and visible to everyone.

Contact us