- #1
franciobr
- 13
- 0
Hello everyone!
I am trying to understand why the following function does not provide problems to being computed numerically:
∫dx1/(sin(abs(x)^(1/2))) from x=-1 to x=2.
Clearly there is a singularity for x=0 but why does taking the absolute value of x and then taking its square root solve the problem?
I searched for quite a while on the internet and on numerical integration books and was surprised that I couldn't find the answer for that. Apparentlly there is a lack of documentation on singularity removal techniques on the web or I am not searching with the right keywords. Any introductory documentation on the subject is welcome!
For the record I am using MATLAB built-in functions such as quadtx() or integral() to solve it but I that's not the point since it turns out even the most simple simpson rule algorithm can deal with that integral.
I am trying to understand why the following function does not provide problems to being computed numerically:
∫dx1/(sin(abs(x)^(1/2))) from x=-1 to x=2.
Clearly there is a singularity for x=0 but why does taking the absolute value of x and then taking its square root solve the problem?
I searched for quite a while on the internet and on numerical integration books and was surprised that I couldn't find the answer for that. Apparentlly there is a lack of documentation on singularity removal techniques on the web or I am not searching with the right keywords. Any introductory documentation on the subject is welcome!
For the record I am using MATLAB built-in functions such as quadtx() or integral() to solve it but I that's not the point since it turns out even the most simple simpson rule algorithm can deal with that integral.