Expand dfrac1x²+a² in a power series in dfrac xa
Expand $\dfrac1{x^2+a^2}$ in a power series in $\dfrac xa$ for $0
Explanation of Method
Each piece, integrated termwise, produces a series involving $\arctan$'s Leibniz sum $\pi/4$ and Catalan's constant $G$; remarkably the $G$-terms from the two pieces have opposite sign and cancel exactly, leaving a clean elementary answer.
- Expand $1/(x^2+a^2)$ on each sub-interval and integrate termwise.
- Evaluate $I_1=\int_0^a$ and $I_2=\int_a^{\infty}$ separately, each producing a $\pi/4$ term and a Catalan's-constant term.
- Add $I_1+I_2$ and watch the Catalan's-constant pieces cancel.
Step 1 · The two series
$$ \frac1{x^2+a^2} = \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{a^{2n+2}}\ (0
Step 2 · Evaluate $I_1$
Using $\int_0^ax^{2n}\ln x\,dx=\dfrac{a^{2n+1}}{(2n+1)^2}[(2n+1)\ln a-1]$ (integration by parts): $$ I_1 = \sum_{n=0}^{\infty}\frac{(-1)^n}{a^{2n+2}}\cdot\frac{a^{2n+1}}{(2n+1)^2}[(2n+1)\ln a-1] = \frac{\ln a}a\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1} - \frac1a\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}. $$ With $\sum(-1)^n/(2n+1)=\pi/4$ and $\sum(-1)^n/(2n+1)^2=G$ (Catalan's constant): $$ I_1 = \frac{\pi\ln a}{4a} - \frac Ga. $$
Step 3 · Evaluate $I_2$ and combine
An analogous integration by parts on $\int_a^{\infty}x^{-(2n+2)}\ln x\,dx=\dfrac1{(2n+1)a^{2n+1}}\left[\ln a+\dfrac1{2n+1}\right]$ gives, after summing: $$ I_2 = \frac{\pi\ln a}{4a} + \frac Ga. $$ The Catalan's-constant terms are equal and opposite, so: $$ I(a) = I_1+I_2 = \frac{\pi\ln a}{4a}+\frac{\pi\ln a}{4a} = \frac{\pi\ln a}{2a}. $$
Summary Table
| Property | Value |
|---|---|
| $I_1$ (on $[0,a]$) | $\pi\ln a/(4a) - G/a$ |
| $I_2$ (on $[a,\infty)$) | $\pi\ln a/(4a) + G/a$ |
| $I(a)$ | $\pi\ln a/(2a)$ |
Additional Notes
- The exact cancellation of Catalan's constant between the two pieces can be verified independently via the substitution $x=at$: it collapses $I(a)$ to $\frac{\ln a}a\int_0^{\infty}\frac{dt}{1+t^2}+\frac1a\int_0^{\infty}\frac{\ln t}{1+t^2}dt$, and the second integral vanishes by the symmetry $t\mapsto1/t$ — the same cancellation, seen from a different angle.
- At $a=1$: $I(1)=0$, consistent with $\int_0^{\infty}\ln(x)/(1+x^2)dx=0$, itself a direct consequence of the $t\mapsto1/t$ antisymmetry noted above.
- This problem is a nice illustration of how splitting an integral at a natural point and expanding separately can produce individually messy (Catalan-constant-laden) pieces that combine into an elementary closed form — the mess is only apparent, not essential.