Prove the relation $\frac{1}{x}$=$\int^\infty_0$ $e^{-xt}$ dt, for $x>0$. Use it to prove $\int^\infty_0$ $\frac{\sin(x)}{x}$ dx = $\frac{\pi}{2}$

Question

Prove the relation $$\frac{1}{x} = \int^\infty_0 e^{-xt}\, \text{d}t, \text{ for } x>0.$$

Use it to prove $$\int^\infty_0\frac{\sin(x)}{x}\, \text{d}x = \frac{\pi}{2}.$$

"Hint: Use appropriate double integrals as in calculus. When the double integrals are proper, the conditions for switching the order of integration are valid. Then take the limits." I'm assuming that's for the second part but I could be wrong.

Answer 1

We have

$$\int_0^\infty\frac{\sin x}{x}dx=\int_0^\infty\int_0^\infty e^{-xt}\sin (x)dtdx=\int_0^\infty\operatorname{Im}\int_0^\infty e^{x(i-t)}dxdt\\=\int_0^\infty\frac1{1+t^2}dt=\arctan t\Bigg|_0^{\infty}=\frac\pi2$$

Answer 2

$$\int_0^\infty e^{-xt}dt=\left[-\frac{1}{x}e^{-xt}\right]_0^\infty =\frac{1}{x}-\underbrace{\lim_{t\to\infty }\frac{1}{x}e^{-xt}}_{=0}=\frac{1}{x}$$