1.14. There is a deeper supplement to the inversion formula (4) or Exercise] 0 above, due to B Rosen TInder the condition J (1 + log Ixl)dF(x) < 00, -00 the improper Riemann integral in Exercise 1 D may be replaced by a Lebesgue integral. [HINT. It is a matter of proving the existence of the latter. Since 1 00 rN I sin(x -y)t I 1 00 -00 dF(y) Jo t dt < -00 dF(y){1 + logO + Nix -yl)} < 00, we have 100 iN sin(x -y)t iN dt 1 00 dF(y) dt = -sin(x -y)t dF(y).

For fixed x, we have