Calculate the following improper integral: integral

Calculate the following improper integral: $$ \int_0^{\infty} dx \int_{2x}^{\infty} x\,e^{-y}\,\frac{\sin y}{y^2}\,dy. $$

Step 1 · Identify and reverse the region

The region is $D=\{(x,y):x\geqslant0,\ y\geqslant2x\}=\{(x,y):y\geqslant0,\ 0\leqslant x\leqslant\tfrac y2\}$, so:

$$ I = \int_0^{\infty} \int_0^{y/2} x\,e^{-y}\,\frac{\sin y}{y^2}\,dx\,dy. $$

Step 2 · Evaluate the inner integral

For fixed $y$, the factor $e^{-y}\sin y/y^2$ is constant with respect to $x$:

$$ \int_0^{y/2} x\,e^{-y}\frac{\sin y}{y^2}\,dx = e^{-y}\frac{\sin y}{y^2}\left[\frac{x^2}2\right]_0^{y/2} = e^{-y}\frac{\sin y}{y^2}\cdot\frac{y^2}8 = \frac18\,e^{-y}\sin y. $$

Step 3 · Evaluate the outer integral

$$ I = \frac18\int_0^{\infty} e^{-y}\sin y\,dy. $$

Using the standard Laplace-transform value $\displaystyle\int_0^\infty e^{-ay}\sin(by)\,dy=\dfrac{b}{a^2+b^2}$ with $a=b=1$:

$$ \int_0^{\infty} e^{-y}\sin y\,dy = \frac{1}{1^2+1^2}=\frac12. $$

(This can also be recovered by integrating by parts twice and solving the resulting linear equation for the integral, as is standard.)

Step 4 · Combine

$$ I = \frac18\cdot\frac12 = \frac1{16}. $$

Additional Notes

• Verified numerically: direct evaluation of the original iterated integral (integrating $y$ over $[2x,\infty)$ then $x$ over $[0,\infty)$) gives $I\approx0.0625000$, matching $\frac1{16}=0.0625$ exactly.