Trying to calculate an integral

  • Mathematica
  • Thread starter Mr Davis 97
  • Start date
  • #1
1,456
44

Main Question or Discussion Point

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.
 

Answers and Replies

  • #2
tnich
Homework Helper
1,048
336
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).
 
  • #3
1,456
44
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.
 
  • #4
tnich
Homework Helper
1,048
336
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.
 
  • #5
tnich
Homework Helper
1,048
336
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.
 
  • #6
tnich
Homework Helper
1,048
336
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.
 
  • #7
lurflurf
Homework Helper
2,426
126
^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.
 
  • #8
203
83
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
 
  • #9
tnich
Homework Helper
1,048
336
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
203
83
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
lurflurf
Homework Helper
2,426
126
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!##
 

Related Threads on Trying to calculate an integral

  • Last Post
Replies
6
Views
1K
Replies
1
Views
1K
Replies
1
Views
784
Replies
14
Views
7K
Replies
1
Views
5K
  • Last Post
Replies
5
Views
3K
Replies
0
Views
845
Replies
2
Views
1K
Top