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 $\displaystyle\int_1^x\frac{\arctan t}t\,dt$ in a power series in $x$ and indicate when the expansion is valid.
Explanation of Method
Dividing the $\arctan t$ series by $t$ gives an even power series valid for $|t|\le1$; integrating from $1$ to $x$ (staying inside this interval) produces a power series in $x$ plus a constant built from Catalan's constant.
- Divide the $\arctan t$ series by $t$.
- Integrate from $1$ to $x$ term by term.
- Identify the resulting constant as Catalan's constant and determine validity.
Step 1 · Divide by $t$
$$ \frac{\arctan t}t = \frac1t\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}t^{2n+1} = \sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}t^{2n}, \qquad |t|\le1. $$
Step 2 · Integrate from 1 to $x$
$$ \int_1^x\frac{\arctan t}t\,dt = \sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\int_1^xt^{2n}\,dt = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}\bigl(x^{2n+1}-1\bigr). $$
Step 3 · Separate the constant (Catalan's constant)
$$ \int_1^x\frac{\arctan t}t\,dt = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}x^{2n+1} - G, \qquad G=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}\approx0.9160 \text{ (Catalan's constant)}. $$ Numerical check at $x=0.5$: the series gives $\approx0.4872-0.9160=-0.4288$, matching a direct numerical (Simpson's rule) estimate of $\int_1^{0.5}\arctan(t)/t\,dt\approx-0.4287$.
Summary Table
| Property | Value |
|---|---|
| Series | $\sum(-1)^n x^{2n+1}/(2n+1)^2-G$ |
| Constant | $G$ = Catalan's constant |
| Region of validity | $|x|\le1$ |
Additional Notes
- At $x=1$ the integral trivially equals $0$ (equal limits), matching $G-G=0$ from the series — a built-in consistency check.
- Catalan's constant $G\approx0.91597$ plays the same structural role here that $\pi^2/12$ played in Problem 264's dilogarithm integral: both arise as the boundary value of a term-by-term-integrated series at the edge of its interval of convergence.
- Since $\arctan(t)/t$ is itself only guaranteed by the series representation for $|t|\le1$, this integral's series representation is intrinsically tied to the interval $[-1,1]$ — extending $x$ beyond $\pm1$ would require a different (non-series) evaluation of the original integral.