Calculate the following integral: double integral over D

Calculate the following integral: $$ \iint_D y\cos(xy^2)\,dx\,dy, $$ where $0<x<\frac{\pi}{2}$ and $0<y<2$.

Step 1 · Identify the region

$D=\left[0,\frac\pi2\right]\times[0,2]$.

Step 2 · Set up the integral

$$ I = \int_0^2\int_0^{\pi/2} y\cos(xy^2)\,dx\,dy. $$

Step 3 · Integrate with respect to $x$

With $u=xy^2$, $du=y^2\,dx$:

$$ y\int_0^{\pi/2}\cos(xy^2)\,dx = \frac1y\int_0^{\pi y^2/2}\cos u\,du = \frac1y\sin\!\left(\frac{\pi y^2}{2}\right). $$

Step 4 · Set up the remaining $y$-integral

$$ I = \int_0^2 \frac{\sin\!\left(\pi y^2/2\right)}{y}\,dy. $$

Step 5 · Substitute to a sine-integral form

Let $w=y^2$, $dw=2y\,dy$, so $\frac{dy}{y}=\frac{dw}{2w}$; when $y=0$, $w=0$; when $y=2$, $w=4$:

$$ I = \int_0^4 \sin\!\left(\frac{\pi w}{2}\right)\frac{dw}{2w}. $$

Now let $t=\frac{\pi w}{2}$, so $w=\frac{2t}{\pi}$, $dw=\frac{2}{\pi}\,dt$; when $w=0$, $t=0$; when $w=4$, $t=2\pi$:

$$ I = \int_0^{2\pi} \sin t \cdot \frac{dt}{2t} = \frac12\int_0^{2\pi}\frac{\sin t}{t}\,dt = \frac12\,\mathrm{Si}(2\pi). $$

Step 6 · Numerical value

$\mathrm{Si}(2\pi)\approx1.4182$, so:

$$ I \approx \frac{1.4182}{2} \approx 0.7091. $$

Additional Notes

• The sine integral $\mathrm{Si}(x)=\int_0^x\frac{\sin t}{t}\,dt$ is a well-tabulated special function, not expressible in elementary terms — its appearance here is expected once the $y$-integral reduces to $\int\frac{\sin(\cdot)}{(\cdot)}\,d(\cdot)$.