- #1
SqueeSpleen
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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.
I have to proof
[tex]I_{n} = I_{n-1} (n-1)/(n+2) \forall n \geq 2[/tex]
n is a natural number.
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
[tex]\int_{0}^{\infty} x^{2n-1}/(x^2+1)^{n+3} \dx = \int_{0}^{\infty} (u-1)^{n-1}/(u)^{n+3} \du[/tex]
In a way similar to the way you use to create pascal's triangle:
[tex]\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 [/tex]
So
[tex]I_n = I_{n-1} - \int_{0}^{\infty} (u-1)^{n-2}/(u)^{n+3} \du [/tex]
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.
Homework Statement
[tex]I_n = \int_{0}^{\infty} x^{2n-1}/(x^2+1)^{n+3} \dx, n \geq 1[/tex]I have to proof
[tex]I_{n} = I_{n-1} (n-1)/(n+2) \forall n \geq 2[/tex]
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
[tex]\int_{0}^{\infty} x^{2n-1}/(x^2+1)^{n+3} \dx = \int_{0}^{\infty} (u-1)^{n-1}/(u)^{n+3} \du[/tex]
In a way similar to the way you use to create pascal's triangle:
[tex]\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 [/tex]
So
[tex]I_n = I_{n-1} - \int_{0}^{\infty} (u-1)^{n-2}/(u)^{n+3} \du [/tex]
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.
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