積分解説20

有名問題です。

$\tau \rho \iota \alpha$ さんの こちら の記事を見て閃いたので、まずはそちらを見てみることをお勧めします。

[解説]
$\begin{array}{rcl}& & {\int }_0^{\mathrm{\infty }}{\left(\frac{\sin x}{x}\right)}^{2}dx\\ & =& {\int }_0^{\mathrm{\infty }}{\sin }^{2}x{\int }_0^{\mathrm{\infty }}{e}^{-xy}dy{\int }_0^{\mathrm{\infty }}{e}^{-xz}dzdx\\ & =& {\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{e}^{-(y+z)x}{\sin }^{2}xdxdydz\\ & =& \frac{1}{2}{\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}(1-\cos 2x)e^{-(y+z)x}dxdydz\\ & =& \frac{1}{2}{\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}({\int }_0^{\mathrm{\infty }}{e}^{-(y+z)x}dx-{\int }_0^{\mathrm{\infty }}\cos 2x{e}^{-(y+z)x}dx)dydz\\ & =& \frac{1}{2}{\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}({[-\frac{1}{y+z}{e}^{-(y+z)x}]}_0^{\mathrm{\infty }}-\mathrm{Re}{\int }_0^{\mathrm{\infty }}{e}^{2ix}{e}^{-(y+z)x}dx)dydz\\ & =& \frac{1}{2}{\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}(\frac{1}{y+z}-\mathrm{Re}{\int }_0^{\mathrm{\infty }}{e}^{-(y+z-2i)x}dx)dydz\\ & =& \frac{1}{2}{\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}(\frac{1}{y+z}-\mathrm{Re}{[-\frac{1}{y+z-2i}{e}^{-(y+z-2i)x}]}_0^{\mathrm{\infty }})dydz\\ & =& \frac{1}{2}{\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}(\frac{1}{y+z}-\mathrm{Re}\frac{1}{y+z-2i})dydz\\ & =& \frac{1}{2}{\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}(\frac{1}{y+z}-\mathrm{Re}\frac{y+z+2i}{(y+z{)}^2+4})dydz\\ & =& \frac{1}{2}{\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}(\frac{1}{y+z}-\frac{y+z}{(y+z{)}^2+4})dydz\\ & =& \frac{1}{2}{\int }_0^{\mathrm{\infty }}{\int }_z^{\mathrm{\infty }}(\frac{1}{y}-\frac{y}{y^2+4})dydz\\ & =& \frac{1}{2}{\int }_0^{\mathrm{\infty }}{[\log y-\frac{1}{2}\log (y^2+4)]}_z^{\mathrm{\infty }}dz\\ & =& \frac{1}{4}{\int }_0^{\mathrm{\infty }}\log (1+\frac{4}{z^2})dz\\ & =& \frac{1}{4}{[z\log (1+\frac{4}{z^2})]}_0^{\mathrm{\infty }}+\frac{1}{4}{\int }_0^{\mathrm{\infty }}z\frac{\frac{8}{z^3}}{1+\frac{4}{z^2}}dz\\ & =& 2{\int }_0^{\mathrm{\infty }}\frac{1}{z^2+4}dz\\ & =& 2{\int }_0^{\frac{\pi }{2}}\frac{1}{2}dz\\ & =& \frac{\pi }{2}\end{array}$

よって、この問題の解答は${\displaystyle \frac{\pi }{2}}$となります。