Show $\int_{s}^{\infty} f(x)dx = \mathcal{L} \{\frac{F(t)}{t}\}$ given $f(x) = \int_{0}^{\infty} e^{-xt}F(t)dt$

Question

I'm trying to derive this to show that

$$\int_{0}^{\infty} f(x)dx = \int_{0}^{\infty} \frac{F(t)}{t} dt$$

and use that to prove

$$\int_{0}^{\infty} \frac{\sin t}{t} = \frac{\pi}{2}$$

How do I go about proving that $$\int_{s}^{\infty} f(x)dx = \mathcal{L} \{\frac{F(t)}{t}\}$$ given $$f(x) = \int_{0}^{\infty} e^{-xt}F(t)dt$$

Answer 1

Observe that $$\frac{e^{-st}}{t} = \int_s^{\infty} e^{-xt} dx.$$ Then

$$\int_0^{\infty} e^{-st} \frac{F(t)}{t} dt = \int_0^{\infty} \int_s^{\infty} e^{-xt} dx F(t) dt = \int_s^{\infty} \int_0^{\infty} e^{-xt} F(t) dt dx = \int_s^{\infty}f(x)dx, $$ where in the second equality I have assumed that $F$ is nice enough so that you can exchange the order of integration.