Question

How do you integrate `f(x) = 2x+4x^3+6x^5+...+2nx^(2n-1)+...` ?

Answer 1

`int f(x) dx = 1/(1-x^2) + C` for `x in (-1, 1)`

Explanation:

For finite sums we know that the integral of the sum is equal to the sum of the integrals.

This extends to infinite sums, provided we know that the sums converge.

`f(x) = lim_(n->oo) [2x+4x^3+6x^5+...+2nx^(2n-1)]=sum_(n=1)^oo 2nx^(2n-1)`

is a power series which will converge for `abs(x) < 1`

`color(white)()`
We can integrate term by term to find:

`int f(x) dx = sum_(n=1)^oo color(red)(cancel(color(black)(2n))) x^(2n) / color(red)(cancel(color(black)(2n))) + C_1 = 1 + sum_(n=1)^oo x^(2n) + C = sum_(n=0)^oo x^(2n) + C = sum_(n=0)^oo (x^2)^n + C = 1/(1-x^2) + C`

The slight trick here is that the constant of integration can be adjusted to give us a familiar sum.

`sum_(n=0)^oo t^n = 1/(1-t)`

with `t = x^2`