Question

How to evaluate `int ln(1-x^9)dx` as a power series?

Answer 1

`int \ ln(1-x^9) \ dx = -x^10/10-1/2x^19/19-1/3x^28/28-1/4x^37/37 - ....`

The general term being:

`u_n = -x^(9n+1)/(n(9n+1))`

Explanation:

A standard power series is:

`ln(1+x) = x-1/2x^2+1/3x^3-1/4x^4 + .... \ \ \` for `|x|<1`

From this we can deduce;

`ln(1-x) = ln(1+(-x) )`
`" " = -x-1/2x^2-1/3x^3-1/4x^4 - .... \ \ \`

And so:

`ln(1-x^9) = -x^9-1/2x^18-1/3x^27-1/4x^36 - ....`

`:. int \ ln(1-x^9) \ dx= int \ -x^9-1/2x^18-1/3x^27-1/4x^36 - .... dx`
`:. " " = -x^10/10-1/2x^19/19-1/3x^28/28-1/4x^37/37 - ....`

The general term being:

`u_n = -x^(9n+1)/(n(9n+1))`

Answer 2

`1/(1-x)=sum_(n=0)^oox^n`

`int1/(1-x)dx=sum_(n=0)^oointx^ndx`

`-ln(1-x)=C+sum_(n=0)^oox^(n+1)/(n+1)`

`x=0` shows that `C=0`:

`ln(1-x)=-sum_(n=0)^oox^(n+1)/(n+1)`

`ln(1-x^9)=-sum_(n=0)^oo(x^9)^(n+1)/(n+1)`

`color(white)(ln(1-x^9))=-sum_(n=0)^oox^(9n+9)/(n+1)`

`intln(1-x^9)dx=-sum_(n=0)^oointx^(9n+9)/(n+1)dx`

`color(white)(intln(1-x^9)dx)=C-sum_(n=0)^oox^(9n+10)/((9n+10)(n+1))`

`x=0` shows `C=0`:

`intln(1-x^9)dx=-sum_(n=0)^oox^(9n+10)/((9n+10)(n+1))`