# Trying to calculate an integral

• Mathematica
• Mr Davis 97

#### Mr Davis 97

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.

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).

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.

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.

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.

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.

• tnich
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 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.

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. Peace,
Fred

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!##