# Finite-part integrals (Hadamard) integrals with Mathematica

• Mathematica
• traianus
In summary: Riemann Summation Operator...That is not true, you need the Ramanujan Summation Operator. It is complete nonsense without that.I had a curious idea..let be a function f(x) with a singularity at x=1 then we define the function:f(x)=f*(x)+\delta(x-1) and by definition f*(1)=0 f(x)=f*(x) for all x except x=1as you can see both function diverge at x=1 however using this definiton:
traianus

Hello guys,
I have to calculate an integral defined in the Hadamard sense with MATHEMATICA, but I only found the option for the integrals defined in the CAUCHY sense.
Just to be clear, suppose to have the integral with limits -1 and +1 of (1/x^2). This integral does not exist in the usual sense, but in the Hadamard sense its value is -2. Now, how do you calculate (numerically) the Hadamard integrals if the functions are complicated using MATHEMATICA?

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So? Nobody knows Hadamard finite-part integrals? Is it so difficult for everybody?

The concept of "Hadamard fnite integral" seems very curious to me could oyo provide a link where to find some info?.. perhaps you should use zeta regularization when calculating your integrals so:

$$\sum_{n=1}^{\infty}n^{s}=\zeta (-s)$$

Hello Karlisbad, I know a very good article: "Numerical evaluation of ypersingular integrals" by Giovanni Monegato, Journal of Computational and Applied Mathematics 50 (1994) 9-31.

I do not know how the zeta regularization works. Can you please explain? Thank you!

A good intro for undergraduates can be found at ..

http://math.ucr.edu/home/baez/twf_ascii/week126

Based on an Euler converntion about divergent series..

resumming let be the divergent series..

$$1+2^{s}+3^{s}+...... \rightarrow \zeta (-s)$$ (1)

of course if s<0 then the series above makes sense (no pole at s=1) and is just the Riemann zeta function.. the algorithm of "Zeta regularization" makes some kind of analytic continuation (i'm not mathematician so i can't give a rigorous formulation by the way really hate math rigour.. ) and then the sum of the series (divergent) can be obtained (¿¿cheating??...if you believe this is cheating you should study renormalization in physics) as the negative values of the Riemann zeta function, for example if s=1 then..

$$1+2+3+......=-1/12$$

if s is an even number then the "sum" is taken to be 0.

further info: "H.G Hardy divergent series" (you have it on E-mule but you need to download an .djvu viewer)

$$1+2+3+......=-1/12$$

That is not true, you need the Ramanujan Summation Operator. It is complete nonsense without that.

I had a curious idea..let be a function f(x) with a singularity at x=1 then we define the function:

$$f(x)=f*(x)+\delta(x-1)$$ and by definition $$f*(1)=0$$

$$f(x)=f*(x)$$ for all x except x=1

as you can see both function diverge at x=1 however using this definiton:

$$\int_{0}^{2}dxf(x)=\int_{0}^{2}dx(f*(x)+\delta(x-1))=1+\int_{0}^{2}dxf*(x)$$ which is now finite...

which is now finite...
It's only finite if f and $\delta$ have a singularity of the same order, and whose coefficients are additive inverses.

It's more typical to use analytically defined functions (like 1/x) to subtract off singularities of functions.

Sigh. Jose might hate the rigour of mathematics, but without it you just get apparent nonsense like this thread. (it should be s<-1, for instance, in post 5, not s<0), not to mention setting out in a confusing manner material which makes it appear that the sum 1+2+3+... actually does exist in the normal sense of summation. Still, never mind, eh? Is * supposed to be convolution? Those aren't functions, by the way. Why is that integral of that 'function' 'now finite'? What if I let f be some truly horrendous function with an essential singularity (infinitely many negative powers in the Laurent expansion)?

If it helps, we don't object to 'non-rigorous' stuff at all; it just isn't maths in the rigorous formal proof sense. We do object to ill defined terms, confusing notation, indecipherable descriptions, and things that are so wrong it is impossible to know where to start to correct them. There is a difference between being rigorous and doing something clearly. No physicist would accept what you've written above as acceptable presentation either. It's nothing to do with mathematicians.

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- The problem with rigour is as perhaps user "eljose" has pointed before in his post is that you can put 'obstacles' to science, the same happened to infinitesimals $$dx$$ , the Dirac delta function $$\delta (x)$$ or "Feynmann (??) Path integrals" in QM or QFT (Field theory) to quantizy everthing..which can't even be defined (rigorously) in terms of Riemann sums¡¡ (sigh), i read a paper by A.Connes about "renormalization and Hopf Algebras" of course the paper were so much rigorous and math-defined that you could hardly "extract" some info about it regarding renormalization, whereas there're lots and lots of math introduction without rigour in physics, conjectures or hypothesis that can be understood by almos everybody..that's why i critizy rigour..i don't hate mathematician i only say that sometimes they are a bit of "pedantry".

Who's putting obstacles there? Physicisist don't require the same level rigour that mathematicians do, so what? Point out one example where rigour held back the development of Feynman path integrals, please.

You aren't trained to read mathematics research papers; they aren't written with people with undergrad physics degrees in mind. They are written by professionals for professionals.

Unrigorous work often helps, and mathematicians use it all the time. Your problem is not that mathematicians are pedantic about rigor. Your problems with getting your work read have been catalogued at length many times; you have had far more peer review than anyone else I can think of; you have yet to address more than one comment (to include proper references, occasionally).

Pointing out that saying 'let f be a function with singularity at 1' is useless is not pedantry (that would be pointing out the grammatical errors, which is not helpful). There are infinitely many such functions. Which one? It is important because what kind of singularity you have affects what follows in your post, making it just plain wrong. And no, you can't get away with saying 'oh, choose one that makes it work'. You don't appear to make any effort to avoid mistakes, so we have no idea what is 'wrong but just a typo' and what is 'just plain wrong'.

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## 1. What is a finite-part integral?

A finite-part integral is a type of integral that is used in mathematics and physics to evaluate functions that are not integrable by traditional methods. It is defined as the limit of a regularized integral where the singularity is removed.

## 2. What is the Hadamard finite-part integral?

The Hadamard finite-part integral is a specific type of finite-part integral that is used in complex analysis. It is named after the French mathematician Jacques Hadamard and is used to evaluate functions with simple poles at the integration limits.

## 3. How is Mathematica used to compute finite-part integrals?

Mathematica is a popular software program used by scientists and mathematicians to perform complex calculations and solve equations. It has built-in functions that allow for the computation of finite-part integrals, including the Hadamard finite-part integral.

## 4. What are the applications of finite-part integrals?

Finite-part integrals have numerous applications in mathematics, physics, and engineering. They are commonly used in the study of complex functions, quantum field theory, and the evaluation of divergent integrals in physical problems.

## 5. Are there any limitations to using finite-part integrals?

While finite-part integrals are a powerful tool for evaluating complex functions, they do have some limitations. They are only applicable to functions with singularities at the integration limits and are not suitable for integrals with branch cuts or multiple poles.

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