Mathematica Trying to calculate an integral

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The discussion revolves around computing the integral of a specific function using Mathematica. The original attempt to calculate the integral resulted in errors, prompting suggestions to adjust the syntax and placement of variables within the sum and product. Participants debated the convergence of the integral, with some recommending analytical methods over numerical ones. Ultimately, they arrived at a polynomial form for the integrand, leading to differing numerical results. The conversation highlights the importance of proper syntax and understanding of integration techniques in Mathematica.
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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