Trying to calculate an integral

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Discussion Overview

The discussion revolves around calculating a specific integral using Mathematica. Participants explore various approaches to evaluate the integral, which involves products and sums of terms dependent on the variable x, over the interval from 0 to 1. The conversation includes attempts at numerical and analytical methods, as well as considerations of convergence.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant seeks assistance in computing the integral using Mathematica, expressing difficulty with the syntax.
  • Another participant points out that the limits for the sum and product were not correctly specified initially.
  • A participant suggests placing the variable x inside the sum to potentially resolve the issue.
  • Concerns are raised about the convergence of the integral, with one participant asserting it does not converge.
  • Another participant counters that the integral is a polynomial and therefore converges.
  • Several alternative integral formulations are proposed, showcasing different ways to express the original problem in Mathematica.
  • One participant advocates for an analytical approach, expanding the integrand and calculating the integral directly, while another mentions discrepancies in numerical results.
  • There is a discussion about the potential for errors in mental calculations of polynomial integrals.
  • Integration by parts or substitution is suggested as a method to avoid expanding the polynomial.
  • A general result involving Pochhammer symbols is introduced, linking it to the integral calculation.

Areas of Agreement / Disagreement

Participants express differing views on the convergence of the integral, with some asserting it converges and others questioning this. There is no consensus on the best approach to solve the integral, as various methods are proposed and debated.

Contextual Notes

Participants mention potential errors in numerical calculations and the complexity of the integral, indicating that assumptions about convergence and the appropriateness of numerical versus analytical methods are not fully resolved.

Mr Davis 97
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This is actually a WolframAlpha question, but I suppose someone conversant in mathematica could give me an answer. How in Mathematica could I compute ##\displaystyle \int_0^1 \left( \prod_{r=1}^3 (x+r)\right) \left(1+x \sum_{r=1}^3 \frac{1}{x+r} \right) ~ dx##. I tried int (Product[x+r, {r, 1}])*(1+ x*(Sum[1/(x+r), {r, 1}])) dx from 0 to 1, but that just gives an errror.
 
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Mr Davis 97 said:
This is actually a WolframAlpha question, but I suppose someone conversant in mathematica could give me an answer. How in Mathematica could I compute ##\displaystyle \int_0^1 \left( \prod_{r=1}^3 (x+r)\right) \left(1+x \sum_{r=1}^3 \frac{1}{x+r} \right) ~ dx##. I tried int (Product[x+r, {r, 1}])*(1+ x*(Sum[1/(x+r), {r, 1}])) dx from 0 to 1, but that just gives an errror.
Looks like you have only specified the lower limits of the sum and product (1), but no the upper limits (3).
 
tnich said:
Looks like you have only specified the lower limits of the sum and product (1), but no the upper limits (3).
Oh. I meant int (Product[x+r, {r, 3}])*(1+ x*(Sum[1/(x+r), {r, 3}])) dx from 0 to 1. But this doesn't work either... Everything works when I remove that x right by the sum, but doesn't work when I add it in.
 
Mr Davis 97 said:
Oh. I meant int (Product[x+r, {r, 3}])*(1+ x*(Sum[1/(x+r), {r, 3}])) dx from 0 to 1. But this doesn't work either... Everything works when I remove that x right by the sum, but doesn't work when I add it in.
Try putting the ##x## inside the sum.
 
tnich said:
Try putting the ##x## inside the sum.
It looks to me like the integral doesn't converge. Try doing the integral analytically (using the calculus) instead of numerically (using Mathematica) and see what you get.
 
tnich said:
It looks to me like the integral doesn't converge. Try doing the integral analytically (using the calculus) instead of numerically (using Mathematica) and see what you get.
Now I'm not sure that it doesn't converge, but still, multiply out the terms in the integral, do the integration, and see what you get. Then you will know if your are getting the right answer from Mathematica.
 
^It is a polynomial it converges.

Try one of these

Integrate[x (x + 1) (x + 2) (x + 3) (1/x + 1/(x + 1) + 1/(x + 2) + 1/(x + 3)), {x, 0, 1}]
Integrate[Product[x + y, {y, 3}] (1 + x Sum[1/(x + z), {z, 3}]), {x, 0, 1}]
Integrate[Sum[Product[x + y, {y, 0, 3}]/(x + z), {z, 0, 3}], {x, 0, 1}]
Integrate[Product[x + y, {y, 3}] Sum[x/(x + z), {z, 0, 3}], {x, 0, 1}]
Integrate[Sum[Pochhammer[x, 4]/(x + z), {z, 0, 3}], {x, 0, 1}]
Does anyone know a shorter way than five?

int (Product[x + r, {r, 1}])*(1 + x*(Sum[1/(x + r), {r, 1}])) dx
Integrate[Product[x + r, {r, 3}] (1 + x*(Sum[1/(x + r), {r, 3}])), {x, 0, 1}]
is the closest Mathematica form to what you wrote Wolframalpha has less strict syntax.
 
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I don't understand why you need a numerical approach. Expand your integrand:
$$\int_{0}^1( 4x^3+18x^2 + 22x +6)dx=20$$

Peace,
Fred
 
Fred Wright said:
I don't understand why you need a numerical approach. Expand your integrand:
$$\int_{0}^1( 4x^3+18x^2 + 22x +6)dx=20$$

Peace,
Fred
I get the same integrand, but 24 for the numerical result.
 
  • #10
tnich said:
I get the same integrand, but 24 for the numerical result.
Your right--my bad. I'm getting too old to integrate polynomials in my head.:sorry:

Peace,
Fred
 
  • #11
Fred Wright said:
I don't understand why you need a numerical approach. Expand your integrand:
$$\int_{0}^1( 4x^3+18x^2 + 22x +6)dx=20$$

Peace,
Fred
You used a numerical approach as 20 is a (wrong) number.
Integration by parts or substitution can be used to avoid expanding the polynomial.
That reminds me of a nice thing
general result
Differences[Pochhammer[{0, 1}, n]]
Integrate[D[Pochhammer[x, n], x], {x, 0, 1}]
##\int_0^1\mathrm{d}(x)_n=(1)_n-(0)_n=n!##
specific n=4
Differences[Pochhammer[{0, 1}, 4]]
Integrate[D[Pochhammer[x, n], x], {x, 0, 1}]
##\int_0^1\mathrm{d}(x)_4=(1)_4-(0)_4=4!##
 

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