# Thread: Integration by reduction formulae

1. ## Integration by reduction formulae

Given that $\displaystyle I_n =$$\displaystyle \int_{0}^{1}\frac {1}{(1+x^2)^n} dx , show that, for n≥2, \displaystyle 2(n-1)I_n = 2^{1-n} + (2n-3)I_{n-1} Please help me to solve this. 2. Originally Posted by geton Given that \displaystyle I_n =$$\displaystyle \int_{0}^{1}\frac {1}{(1+x^2)^n} dx$, show that, for n≥2,

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

Its less messy to prove:

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

First step: Integration by parts-

$\displaystyle I_{n} =$$\displaystyle \int_{0}^{1}\frac {1}{(1+x^2)^{n}} \, dx = \frac{x}{(1+x^2)^n}\bigg{|}_0^1 + 2n \int_0^1 \frac{x^2}{(1+x^2)^{n+1}} \, dx Second Step: Partial Fractions Now we observe that \displaystyle \int_0^1 \frac{x^2}{(1+x^2)^{n+1}} \,dx = \int_0^1 \frac{1}{(1+x^2)^n}\, dx - \int_0^1 \frac{1}{(1+x^2)^{n+1}}\, dx = I_{n} - I_{n+1}. Third Step: Rearranging \displaystyle I_{n} = 2^{-n} + 2n(I_{n} - I_{n+1}) \displaystyle 2n I_{n+1} = 2^{-n} + (2n-1)I_{n} 3. Originally Posted by geton Given that \displaystyle I_n =$$\displaystyle \int_{0}^{1}\frac {1}{(1+x^2)^n} dx$, show that, for n≥2,

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

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