- Show that \(\int _ { 0 } ^ { \infty } { \frac { \sin { x } } { x } } d x = { \frac { \pi } { 2 } }\)
[Hint: The integral equals \(\begin{array} { r } { \frac { 1 } { 2 i } \int _ { - \infty } ^ { \infty } \frac { e ^ { i x } - 1 } { x } } \end{array}\) dx. Use the indented semicircle.]
Solution
For \(R>0\), let \(C_R\) be the positively oriented upper semicircle from \(R\) to \(-R\). The function \[ F(z)=\frac{e^{iz}-1}{z} \] has a removable singularity at \(0\), so after defining \(F(0)=i\) it is entire. Integrating around the upper half-disc gives \[ \int_{-R}^R\frac{e^{ix}-1}{x}\,dx +\int_{C_R}\frac{e^{iz}-1}{z}\,dz=0. \tag{1} \]
Split the arc integral. Parametrizing \(C_R\) by \(z=Re^{i\theta}\), \(0\le\theta\le\pi\), gives \[ \int_{C_R}\frac{dz}{z}=i\pi. \tag{2} \] Also \[ \left|\int_{C_R}\frac{e^{iz}}{z}\,dz\right| \le\int_0^\pi e^{-R\sin\theta}\,d\theta. \] Using \(\sin\theta\ge2\theta/\pi\) on \([0,\pi/2]\) and symmetry, \[ \int_0^\pi e^{-R\sin\theta}\,d\theta \le2\int_0^{\pi/2}e^{-2R\theta/\pi}\,d\theta \le\frac\pi R\longrightarrow0. \tag{3} \] Thus (1)–(3) imply \[ \lim_{R\to\infty}\int_{-R}^R\frac{e^{ix}-1}{x}\,dx=i\pi. \tag{4} \] The real part of the integrand, \((\cos x-1)/x\), is odd, while its imaginary part, \(\sin x/x\), is even. Hence \[ \int_{-R}^R\frac{e^{ix}-1}{x}\,dx =2i\int_0^R\frac{\sin x}{x}\,dx. \] Taking \(R\to\infty\) and using (4) gives \[ 2i\int_0^\infty\frac{\sin x}{x}\,dx=i\pi, \] so \[ \boxed{\displaystyle \int_0^\infty\frac{\sin x}{x}\,dx=\frac\pi2}. \]