Solving Integral Recursion: Proving I(n) for n∈ℕ0

[Alexx1]

Can someone help me with this? I don't know how to begin

I(n) = integral 1/(1+x^2)^n dx with n ∈ ℕ0

Than: ∀ n ∈ ℕ0, n≠1 : I(n) = 1/(2(n-1)) * x/((1+x^2)^(n-1)) + (2n-3)/(2(n-1)) * I(n-1)

I have to prove this, but I don't know how to start

 

[Petar Mali]

Learn latex for beginning

$$I_n=\int \frac{dx}{(1+x^2)^n}$$

Try to do this integral for $$n=0,2,3...$$

 

[arildno]

Hint:
Write:
$$I_{n}=\int{1}*\frac{1}{(1+x^{2})^{n}}dx$$

 

[Alexx1]

Hint:
Write:
$$I_{n}=\int{1}*\frac{1}{(1+x^{2})^{n}}dx$$

I know, than I can do:

u = 1/((1+x^2)^n) --> du = (-2nx)/((1+x^2)^(n+1))s
dv = 1 --> v = x

= x/((1+x^2)^n) + 2n integral x^2/((1+x^2)^(n+1))

But what now?

 

[arildno]

Rewrite:
$$\frac{x^{2}}{(1+x^{2})^{n+1}}=\frac{x^{2}+1-1}{(1+x^{2})^{n+1}}=\frac{1}{(1+x^{2})^{n}}-\frac{1}{(1+x^{2})^{n+1}}$$
See if that helps..

 

[Alexx1]

Rewrite:
$$\frac{x^{2}}{(1+x^{2})^{n+1}}=\frac{x^{2}+1-1}{(1+x^{2})^{n+1}}=\frac{1}{(1+x^{2})^{n}}-\frac{1}{(1+x^{2})^{n+1}}$$
See if that helps..

For the moment I have:

I(n) = x/((1-2n)((1+x^2)^n)) - (2n/(1-2n))*I(n+1)

What do I have to do next?

 

[arildno]

Solve that equation for I(n+1) instead, and report back!

 

[Alexx1]

Solve that equation for I(n+1) instead, and report back!

I(n+1) = x/((x^2+1)^(n+1)) + 2(n+1)*(I(n+1)-I(n+2))

 

[arildno]

No!
Just rearrange, and get:

$$I_{n+1}=\frac{1}{2n}\frac{x}{(1+x^{2})^{n}}+\frac{2n-1}{2n}I_{n}$$
Verify this, and then see if you manage the last step in the derivation on your own.

 

[Alexx1]

No!
Just rearrange, and get:

$$I_{n+1}=\frac{1}{2n}\frac{x}{(1+x^{2})^{n}}+\frac{2n-1}{2n}I_{n}$$
Verify this, and then see if you manage the last step in the derivation on your own.

Ok, now I see it, you have to change n into n-1 and than you get the result
Thank you very much!

 

[arildno]

Ok, now I see it, you have to change n into n-1 and than you get the result

Right!
Note that in our derivation n=0 is prohibited, and by the index change this entails prohibition on n=1 instead

Thank you very much!

My pleasure!

 

[Alexx1]

Right!
Note that in our derivation n=0 is prohibited, and by the index change this entails prohibition on n=1 instead

My pleasure!

n ∈ ℕ0

with ℕ0 , I meant: 'all the natural numbers except for 0'

 

[arildno]

n ∈ ℕ0

with ℕ0 , I meant: 'all the natural numbers except for 0'

I know.

Note, however, that with just prior to dividing with 2n, we have the equation:

2nI(n+1)=x/(1+x^2)^{n}+(2n-1)I(n)

Note that setting n=0 here yields: 0=x-I(0), which is correct, up to an arbitrary constant C...