Using the power series expansions of e^x, sin x, cos x
Using the power series expansions of $e^x$, $\sin x$, $\cos x$, $\ln(1+x)$, $(1+x)^a$, expand $\displaystyle\int_0^x\frac{\ln(1+t)}t\,dt$ in a power series in $x$ and indicate when the expansion is valid.
Explanation of Method
Dividing the $\ln(1+t)$ series by $t$ removes the singularity at $t=0$ term by term, leaving a series that can be integrated directly; the result is (up to sign) the dilogarithm function.
- Divide the Maclaurin series for $\ln(1+t)$ by $t$.
- Integrate term by term from $0$ to $x$.
- Determine convergence, including both endpoints.
Step 1 · Divide by $t$
$$ \frac{\ln(1+t)}t = \frac1t\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}nt^n = \sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}t^n, \qquad |t|\le1,\ t\ne-1. $$
Step 2 · Integrate term by term
$$ \int_0^x\frac{\ln(1+t)}t\,dt = \sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}\int_0^xt^n\,dt = \sum_{n=0}^{\infty}\frac{(-1)^n}{(n+1)^2}x^{n+1} = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2}x^n. $$ This is the negative dilogarithm: $\int_0^x\frac{\ln(1+t)}t\,dt = -\operatorname{Li}_2(-x)$.
Step 3 · Region of convergence, including endpoints
By the ratio test, $R=1$. At $x=1$: $\sum(-1)^{n-1}/n^2=\eta(2)=\pi^2/12$, converges absolutely. At $x=-1$: $\sum(-1)^{n-1}(-1)^n/n^2=-\sum1/n^2=-\pi^2/6$, converges absolutely. So both endpoints are included.
Summary Table
| Property | Value |
|---|---|
| Series | $\sum(-1)^{n-1}x^n/n^2$ |
| Closed form | $-\operatorname{Li}_2(-x)$ |
| Value at $x=1$ | $\pi^2/12$ |
| Region of validity | $-1\le x\le1$ |
Additional Notes
- Because the $1/n^2$ decay is fast enough for absolute convergence at both endpoints, this series converges on the full closed interval $[-1,1]$, unlike many of the $1/n$-type log series (e.g. Problem 261) which only converge conditionally or not at all at one endpoint.
- The value at $x=1$, $\pi^2/12$, is the Dirichlet eta function $\eta(2)$, related to the Basel-problem constant $\zeta(2)=\pi^2/6$ by $\eta(2)=(1-2^{1-2})\zeta(2)=\tfrac12\zeta(2)$.
- This dilogarithm-type integral is a direct relative of the classical polylogarithm family $\operatorname{Li}_s(x)=\sum x^n/n^s$; taking $s=1$ instead recovers the ordinary logarithm series from Problem 261's building blocks.