Question

Evaluate the indefinite integral as a power series?

`int x^3 ln(1+x)dx`
How would you evaluate this indefinite integral as a power series? And what would the radius of convergence be?

I'm super confused by this topic

Answer 1

`C+sum_(n=1)^oo(-1)^(n-1)x^(n+4)/(n(n+4))` `R=1`

Explanation:

Recall the power series expansion for `ln(1+x):`

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

This is one you should memorize; however, it is derived as follows:

`ln(1+x)=int1/(1+x)dx=int1/(1-(-x))dx`

`=intsum_(n=0)^oo(-1)^nx^n=sum_(n=0)^ooint(-1)^nx^n`

`ln(1+x)=C+sum_(n=0)^oo(-1)^nx^(n+1)/(n+1)` (Term-by-term integration performed on the series)

Letting `x=0,`

`C=ln(1+0)=0`

Performing an index shift to `n=1`, entails replacing all `n` in the series with `n-1`

`=sum_(n=1)^oo(-1)^(n-1)x^n/n`.

Knowing this, we may rewrite our given indefinite integral as follows:

`intx^3sum_(n=1)^oo(-1)^(n-1)x^n/ndx`

Multiply in the `x^3` into the series. We can do this because as far as the series is concerned, `x` will be a fixed value. All we have to do is add `3` to the exponent of `x^n, x^3x^n=x^(n+3)`

`intsum_(n=1)^oo(-1)^(n-1)x^(n+3)/ndx`

The radius of convergence of this series is `R=1,` as that is the radius of convergence of the power series expansion for `ln(1+x)`. Multiplying in the `x^3` does not change the radius of convergence.

We perform term-by-term integration on the series:

`sum_(n=1)^ooint(-1)^(n-1)x^(n+3)/ndx`

`=C+sum_(n=1)^oo(-1)^(n-1)x^(n+4)/(n(n+4))`

We leave `C` as it is here.

The radius of convergence is still `R=1.` The radius of convergence does not change when integrating the series (the interval can).