# Integration by reduction formulae

• May 7th 2008, 09:43 PM
geton
Integration by reduction formulae
Given that $I_n =$ $\int_{0}^{1}\frac {1}{(1+x^2)^n} dx$, show that, for n≥2,

$2(n-1)I_n = 2^{1-n} + (2n-3)I_{n-1}$

• May 7th 2008, 11:14 PM
Isomorphism
Quote:

Originally Posted by geton
Given that $I_n =$ $\int_{0}^{1}\frac {1}{(1+x^2)^n} dx$, show that, for n≥2,

$2(n-1)I_n = 2^{1-n} + (2n-3)I_{n-1}$

Its less messy to prove:

$\boxed{2nI_{n+1} = 2^{-n} + (2n-1)I_{n}}$

First step: Integration by parts-

$I_{n} =$ $\int_{0}^{1}\frac {1}{(1+x^2)^{n}} \, dx = \frac{x}{(1+x^2)^n}\bigg{|}_0^1 + 2n \int_0^1 \frac{x^2}{(1+x^2)^{n+1}} \, dx$

Second Step: Partial Fractions

Now we observe that $\int_0^1 \frac{x^2}{(1+x^2)^{n+1}} \,dx = \int_0^1 \frac{1}{(1+x^2)^n}\, dx - \int_0^1 \frac{1}{(1+x^2)^{n+1}}\, dx = I_{n} - I_{n+1}$.

Third Step: Rearranging

$I_{n} = 2^{-n} + 2n(I_{n} - I_{n+1})$
$2n I_{n+1} = 2^{-n} + (2n-1)I_{n}$
• May 7th 2008, 11:30 PM
mr fantastic
Quote:

Originally Posted by geton
Given that $I_n =$ $\int_{0}^{1}\frac {1}{(1+x^2)^n} dx$, show that, for n≥2,

$2(n-1)I_n = 2^{1-n} + (2n-3)I_{n-1}$

Use integration by parts: $\int u \, dv = uv - \int v \, du$.
Let $u = \frac{1}{(1 + x^2)^n} \Rightarrow du = - \frac{2n x}{(1 + x^2)^{n+1}} \, dx$ and $dv = dx \Rightarrow v = x$.
Note that $\frac{x^2}{(1 + x^2)^{n+1}} = \frac{(1 + x^2) - 1}{(1 + x^2)^{n+1}} = \frac{1}{(1 + x^2)^{n}} - \frac{1}{(1 + x^2)^{n+1}}$.
Then $I_n = \frac{1}{2^n} + 2n (I_n - I_{n+1})$.
Therefore: $2n I_{n+1} = \frac{1}{2^n} + (2n-1) I_n$.
Re-label $n \rightarrow n - 1$ and solve for $I_n$.