By expanding the integrand in a power series, calculate
By expanding the integrand in a power series, calculate $\displaystyle\int_0^{\pi/4}\sin x^2\,dx$ with accuracy $10^{-3}$.
Explanation of Method
Substituting $t=x^2$ into the $\sin t$ series and integrating term by term produces an alternating series in powers of $\pi/4$; because the factorial denominators grow so fast, only a couple of terms are needed.
- Substitute $t=x^2$ into the $\sin t$ series.
- Integrate term by term over $[0,\pi/4]$.
- Evaluate numerically and check the error bound.
Step 1 · Substitute and expand
$$ \sin x^2 = \sum_{n=0}^{\infty}\frac{(-1)^nx^{4n+2}}{(2n+1)!}. $$
Step 2 · Integrate over $[0,\pi/4]$
$$ \int_0^{\pi/4}\sin x^2\,dx = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!(4n+3)}\left(\frac\pi4\right)^{4n+3}. $$
Step 3 · Evaluate numerically
With $\pi/4\approx0.785398$, the needed powers are $(\pi/4)^3\approx0.484473$, $(\pi/4)^7\approx0.184344$, $(\pi/4)^{11}\approx0.070163$: $$ \begin{aligned} n=0:&\ \frac{0.484473}3=0.161491 \\ n=1:&\ -\frac{0.184344}{3!\cdot7}=-0.004389 \\ n=2:&\ \frac{0.070163}{5!\cdot11}=0.000053 \end{aligned} $$ Sum: $0.161491-0.004389+0.000053\approx0.157155$. The next term is of order $10^{-7}$, far below the required $10^{-3}$.
Summary Table
| Property | Value |
|---|---|
| Series | $\sum(-1)^n(\pi/4)^{4n+3}/[(2n+1)!(4n+3)]$ |
| Result | $0.157$ |
| Related function | Fresnel-type integral |
Additional Notes
- This is a close relative of the Fresnel sine integral $S(x)=\int_0^x\sin(\pi t^2/2)\,dt$, important in diffraction and wave optics; both arise from the same "expand, then integrate" treatment of $\sin(\text{quadratic})$.
- Because the argument $\pi/4<1$ and the denominators grow factorially, convergence here is even faster than in the sine-integral case of Problem 284 despite the more complicated $4n+3$ exponent pattern.
- Only three terms were needed to certify $10^{-3}$ accuracy, again illustrating the practical payoff of the remainder bound established in Problem 281.