Integral: recursion

1. Jul 29, 2010

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

2. Jul 29, 2010

Petar Mali

Learn latex for beggining

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

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

3. Jul 29, 2010

arildno

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

Last edited: Jul 29, 2010
4. Jul 29, 2010

Alexx1

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?

5. Jul 29, 2010

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..

6. Jul 29, 2010

Alexx1

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?

7. Jul 29, 2010

arildno

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

8. Jul 29, 2010

Alexx1

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

9. Jul 29, 2010

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.

10. Jul 29, 2010

Alexx1

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

11. Jul 29, 2010

arildno

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

12. Jul 29, 2010

Alexx1

n ∈ ℕ0

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

13. Jul 29, 2010

arildno

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...