Chapter 15.3, Problem 19E

(a)

To determine

To show: The expression is ${\displaystyle {\int }_0^{\infty }\frac{\sin 2x}{x}}dx=\frac{\pi }{2}$ using the equation $f\left(x\right)=\frac{2}{\pi }{\displaystyle {\int }_0^{\infty }\frac{\sin \alpha \cos \alpha \left(x-1\right)}{\alpha }}d\alpha$.

(b)

To determine

To show: The expression is ${\displaystyle {\int }_0^{\infty }\frac{\sin kx}{x}}dx=\frac{\pi }{2}$ for $k>0$.