Question

We have `f:RR->RR,f(x)=1/(1+x^2)` and `I_n=int_0^1f^n(x)dx,ninNN`. Prove that `2nI_(n+1)=(2n-1)I_n+1/(2^n),ninNN`*?

Answer 1

Start with `"I"_n`.

`"I"_n = int_0^1 (1+x^2)^(-n) "d"x`.

Perform a partial integration.

`"I"_n = [x(1+x^2)^(-n)]_0^1-int_0^1 x (-n) (1+x^2)^(-(n+1))(2x) "d"x`
`"I"_n = 1/2^n + 2n \int_0^1 x^2/(1+x^2)^(n+1) "d"x`

At this point, the step taken seems to be a bit random. Observation of the reduction formulae you are trying to prove can often help at this point. Notice that,

`1/(1+x^2)^(n) - 1/(1+x^2)^(n+1) = (1+x^2-1)/(1+x^2)^(n+1)`
`1/(1+x^2)^(n) - 1/(1+x^2)^(n+1) = x^2/(1+x^2)^(n+1)`

Then, substituting,

`"I"_n = 1/2^n + 2n(\int_0^1 1/(1+x^2)^(n) "d"x - int_0^1 1/(1+x^2)^(n+1) "d"x)`,
`"I"_n = 1/2^n + 2n(I_n - I_(n+1))`,
`2nI_(n+1)=(2n-1)I_n+1/(2^n)`.