Solving Integrals: Proofing I_n = I_{n-1} (n-1)/(n+2)

[SqueeSpleen]

I don't speak English very well, so it's very hard to me to explain my attemps to solve this problem, and I'm still learning to use latex, so it's so slow to me. I can scan my attemps if you want to see them.

Homework Statement

I_n = \int_{0}^{\infty} x^{2n-1}/(x^2+1)^{n+3} \dx, n \geq 1
I have to proof
I_{n} = I_{n-1} (n-1)/(n+2) \forall n \geq 2

Homework Equations

n is a natural number.

The Attempt at a Solution

I don't speak English very well, so it's very hard to me to explain my attemps to solve this problem, and I'm still learning to use latex, so it's so slow to me. I can scan my attemps if you want to see them.
The integral's solution is a partcial fractions.
It's a sum from i=0 to n = (n!/((n-i)!(i!)))/(i+3)
(The combinatory number between i and n / i+3).
For example:
For n = 3
1/3-2/4+1/5
For n = 4
1/3-/4+3/5-1/6

Using u=x^2+1
\int_{0}^{\infty} x^{2n-1}/(x^2+1)^{n+3} \dx = \int_{0}^{\infty} (u-1)^{n-1}/(u)^{n+3} \du

In a way similar to the way you use to create pascal's triangle:

\int_{0}^{\infty} (u-1)^{n-1}/(u)^{n+3} \du = \int_{0}^{\infty} (u-1)^{n-2}/(u)^{n+2} - (u-1)^{n-2}/(u)^{n+3} \du

So

I_n = I_{n-1} - \int_{0}^{\infty} (u-1)^{n-2}/(u)^{n+3} \du

For example for n=4 (1/3-3/4+3/5-1/6) = (1/3-2/4+1/5)+(-1/4+2/5-1/6).

I don't write more because I think it's better to use my time to keep trying to solve this (It's SO slow to me write things in latex, and without latex a lot of things are almost impossible to write), I think I prove you I already tried to solve this.

Ah, I don't know how to write dx, du, dy, etc while using LaTeX in the integrals.

 

[tiny-tim]

Welcome to PF!

Hi SqueeSpleen! Welcome to PF!

(Nice LaTeX, btw! )

Ah, I don't know how to write dx, du, dy, etc in the integrals.

When you use substitution in an integral, you must do two things …

i] change the limits (in this case, u goes from 1 to ∞, doesn't it? )

ii] change the dx (in this case, du = 2x dx).

(btw, you must always write the dx or du after an ∫ )

However, in this case, I don't think you need to substitute …

you should be able to get the solution very quicly by using https://www.physicsforums.com/library.php?do=view_item&itemid=199" ​

 

[SqueeSpleen]

I mean, I don't know how to write dx, du, dy, etc with LaTeX in the integrals.
I used integration by parts, I know how to do one integral, but I can't proof the general term.
Thank you, I registered a months ago but I don't post very often because my English is a pretty bad (I have to think a lot to make this posts and I'm almost sure I've done a lot of misstakes here).
I don't know why but I didn't thought I could solve this using integration by parts in the general integral (to every n).
I'll try.
\int_{0}^{\infty} x^{2n-1}/(x^2+1)^{n+3} \dx = (u-1)^{n-1}/((n+2)u^{n+2}) + \int_{0}^{\infty} (n-1)(u-1)^{n-2}/((n+2)(u)^{n+2}) \du
I thought parts wasn't the solution.
But then I realized that
(u-1)^{n-1}/((n+2)u^{n+2})
x^{2n-2}/((n+2)(x^2+1)^{n+2})
Is equal to 0 because it's evaluated between 0 and infinity, and both tend to zero, when x= 0 then is 0 because the upper part (I don't know how do you call it in English), and when x->infinity both parts tend to infinity, but the lower part has an higher exponent, so it tends to zero.
Then:
\int_{0}^{\infty} x^{2n-1}/(x^2+1)^{n+3} \dx = \int_{0}^{\infty} (n-1)(u-1)^{n-2}/((n+2)(u)^{n+2}) \du

\int_{0}^{\infty} x^{2n-1}/(x^2+1)^{n+3} \dx = (n-1)/(n+2) \int_{0}^{\infty} (u-1)^{n-2}/(u)^{n+2} \du

Which is equal to

I_{n} = I_{n-1} (n-1)/(n+2) \forall n \geq 2

The other steps of induction are more easy to do.

Thanks, if you would have solve this instead me I should feel an idiot =/

 

[SqueeSpleen]

Edited.

 

[tiny-tim]

Looks good!

Officially, those parts are called the "numerator" and the "denominator", but nearly everyone actually calls them the "top" and the "bottom".

(btw, you should by now be able to do these without substitution:

remove an x from x^2n-1 and you have x/(x² + 1)ⁿ⁺³,

which you should be able to see immediately is a factor times the derivative of 1/(x² + 1)ⁿ⁺²)

 

[SqueeSpleen]

Yes, but with substitution looks a bit easier to me.
Denominator and numerator :P, I should tried to "translate by instint" it :P, in Spanish they're denominador and numerador.

 

[Char. Limit]

Yes, but with substitution looks a bit easier to me.
Denominator and numerator :P, I should tried to "translate by instint" it :P, in Spanish they're denominador and numerador.

That sort of thing tends to happen when we take our terms from Latin. I've noticed that most if not all math terms in English are cognates with either Latin or German words.

 

[SqueeSpleen]

Oh, when I was writing it in LaTeX I forgot a 1/2 after the substitution.
Anyway it isn't important (in this problem).