Using Simpson's method with n=5, calculate Catalan's

Using Simpson's method with $n=5$, calculate Catalan's constant $G=\displaystyle\int_0^1\frac{\arctan x}{x}\,dx$.

Explanation of Method

The integrand has a removable singularity at $x=0$ (its limit there is $1$, from the Taylor series $\arctan x=x-x^3/3+\cdots$), so defining $f(0)=1$ makes it smooth on the whole interval; Simpson's rule with $2n=10$ subintervals is then applied directly.

Step 1: Handling the singularity at $x=0$

Since $\dfrac{\arctan x}{x}=1-\dfrac{x^2}{3}+\dfrac{x^4}{5}-\cdots$ for small $x$, the limit as $x\to0$ is $1$, so we set $f(0)=1$ to make $f$ continuous (and smooth) on $[0,1]$.

Step 2: Simpson's formula, $2n=10$, $h=0.1$

$x_i$	0	0.1	0.2	0.3	0.4	0.5	0.6	0.7	0.8	0.9	1.0
$f(x_i)$	1.000000	0.996687	0.986978	0.971523	0.951266	0.927295	0.900699	0.872466	0.843426	0.814239	0.785398
weight	1	4	2	4	2	4	2	4	2	4	1

Weighted sum $S\approx27.478971$:

\[G_{\text{Simpson}}=\frac{0.1}{3}\times27.478971\approx0.915966.\]

\[\boxed{G\approx0.915966}\]

Step 3: Comparison with the known value

Catalan's constant is known to be $G\approx0.915965594$, so the actual error here is on the order of $10^{-7}$ — accurate to about 6–7 decimal places from just 10 subintervals, reflecting how smooth and well-behaved $\dfrac{\arctan x}{x}$ is on $[0,1]$.

Step 4: Remark

$G$ also equals the alternating series $\sum_{n=0}^\infty\dfrac{(-1)^n}{(2n+1)^2}=1-\tfrac19+\tfrac1{25}-\cdots$, but that series converges far too slowly for practical high-precision computation, which is why the integral representation combined with a fast-converging rule like Simpson's is the more efficient route.

Result

Final answer:

\[\boxed{G=\int_0^1\frac{\arctan x}{x}\,dx\approx0.915966\ (\text{true value}\approx0.9159656),\ \ E\sim10^{-7}}\]