I updated my work to show the steps of how I got my expansion.
I just want to know if what I have worked on so far is correct or if I messed something up along the way.
I have the function
$$f(x) = \int_0^x \frac {\log(1+t)}{t}dt$$
I would like a confirmation for my Taylor expansion of $\log(1+t)$ about $x_0 = 0$. I got the following,
$$\log(1+t) = \sum_{k=1}^{n} (-1)^{k+1} \;\frac {t^k}{k}+\; \frac {(-1)^n \; t^{n+1}}{(n+1)(1+\xi(t))^{n+1}}$$
Where $\xi$ is between 0 and t. (Below are my steps of how I got this)
First we have that the $n^{th}$ derivative of $log(1+t)$ is
$$\frac{(-1)^{n+1} \; (n-1)!}{(1+t)^n}$$
and therefore for the $n+1$ derivative we have
$$\frac{(-1)^{n} \; n!}{(1+t)^{n+1}}$$
Taylors Theorem with Remainder states $f(x) = p_n(x) + R_n(x)$ for
$$p_n(x)=\sum_{k=0}^{n} \frac {(x-x_0)^k}{k!} \; f^{(k)}(x_0)$$
and
$$R_n(x) = \frac {(x-x_0)^{n+1}}{(n+1)!} \; f^{(n+1)}(\xi_x)$$
for $\xi_x$ between $x_0$ and $x$.
This is what I used to derive the remainder term. (the pointwise version not the integral version).
So I have
$$R_n(t) = \frac {t^{n+1}}{(n+1)!} \; \frac{(-1)^{n} \; n!}{(1+t)^{n+1}} = \frac {(-1)^n \; t^{n+1}}{(n+1)(1+\xi(t))^{n+1}}$$
If this is correct is it ok to just divide by $t$ to get a Taylor expansion for the integrand, and if so what do I have to do to ensure I am handling $t=0$ correctly?
Thank you!!!