By expanding the integrand in a power series, calculate
By expanding the integrand in a power series, calculate $\displaystyle\int_0^1\frac{\sin x}x\,dx$ with accuracy $10^{-5}$.
Explanation of Method
Dividing the $\sin x$ series by $x$ removes the removable singularity at $0$; integrating term by term over $[0,1]$ gives an alternating series whose terms shrink factorially fast, so very few terms are needed for $10^{-5}$ accuracy.
- Divide the $\sin x$ series by $x$.
- Integrate term by term over $[0,1]$.
- Sum terms until the Leibniz bound guarantees $10^{-5}$ accuracy.
Step 1 · Divide by $x$
$$ \frac{\sin x}x = \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n+1)!}. $$
Step 2 · Integrate over $[0,1]$
$$ \int_0^1\frac{\sin x}x\,dx = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)(2n+1)!}. $$
Step 3 · Sum until the error bound is met
$$ \begin{aligned} n=0:&\ 1.000000 \\ n=1:&\ -\tfrac1{18}=-0.055556 \\ n=2:&\ \tfrac1{600}=0.001667 \\ n=3:&\ -\tfrac1{35280}=-0.000028 \\ n=4:&\ \tfrac1{3265920}\approx0.0000003<10^{-5} \end{aligned} $$ Summing $n=0$ to $3$: $$ 1-0.055556+0.001667-0.000028 = 0.946083, $$ with error bounded by the $n=4$ term, $<10^{-5}$.
Summary Table
| Property | Value |
|---|---|
| Series | $\sum(-1)^n/[(2n+1)(2n+1)!]$ |
| Special function | Sine integral, $\operatorname{Si}(1)$ |
| Result | $0.94608$ |
Additional Notes
- This integral is exactly the sine integral function evaluated at $1$, $\operatorname{Si}(1)$; the true value $0.9460830704\ldots$ matches our four-term sum to all six displayed decimals, illustrating how fast factorial-denominator series converge.
- Compare with Problem 266's cosine integral, $\operatorname{Ci}(x)$: both are classical special functions defined precisely because $\sin x/x$ and $\cos x/x$ have no elementary antiderivative.
- Only four terms were needed here versus the hundreds of thousands required for the raw Leibniz series in Problem 282 — a direct consequence of the extra $1/(2n+1)!$ decay contributed by dividing by $x$ and integrating a factorial series.