Cauchy law and its characteristic function

The nonnegative exponential density has integral ${\lim }_{R\to \mathrm{\infty }}{\int }_0^R{e}^{-x}dx={\lim }_R(1-e^{-R})=1$, using the compact primitive, bridge and monotone convergence. For real t the function $e^{(-1+it)x}\mathrm{/}(-1+it)$ is a primitive of $e^{-x}(\cos (tx)+i\sin (tx))$, as direct componentwise differentiation shows; the denominator cannot vanish since its real part is -1. Therefore ${\int }_0^R{e}^{itx}{e}^{-x}dx=\frac{e^{(-1+it)R}-1}{-1+it}⟶\frac{1}{1-it}\mathrm{.}$ The boundary modulus is $e^{-R}$ and the integrand modulus is $e^{-x}$, so DCT justifies the full oscillatory integral. Whenever density integration is used below for a bounded complex test $u+iv$, apply [F16] separately to the nonnegative functions $u^{+},u^{-},v^{+},v^{-}$ and then reassemble their finite integrals componentwise as in [F1]; this gives $\int (u+iv)d(fdx)=\int (u+iv)fdx$ without enlarging [F16]. Reflection substitution on compact intervals and then DCT show that the reflected density has transform $1\mathrm{/}(1+it)$. Splitting the two half-lines gives the probability density $\mathrm{\ell }(x)=e^{-\mathrm{\mid }x\mathrm{\mid }}\mathrm{/}2$ and transform $\frac{1}{2}[(1-it{)}^{-1}+(1+it{)}^{-1}]=(1+t^2{)}^{-1}$.

The increasing arctangent has range $(-\pi \mathrm{/}2,\pi \mathrm{/}2)$, so its limits at the two infinities are the endpoints of that range: any smaller limiting supremum would omit values in the range, and similarly for the infimum. Consequently compact arctangent integration and monotone convergence give ${\int }_{\mathbb{R}}(1+t^2{)}^{-1}dt=\pi$. The Laplace characteristic function is therefore integrable. Density inversion supplies the continuous density $h(x)=(2\pi {)}^{-1}\int e^{-itx}\mathrm{/}(1+t^2)dt$ for the same Laplace law. It equals $\mathrm{\ell }$ at every point: if two continuous densities of the same measure differed at one point, their difference would have a fixed strict sign and magnitude on a small interval, contradicting equal integrals on that interval. Hence $\frac{1}{\pi }{\int }_{\mathbb{R}}\frac{e^{itx}}{1+x^2}dx=2h(-t)=e^{-\mathrm{\mid }t\mathrm{\mid }}\mathrm{.}$ The preceding arctangent integral also normalizes c, and density integration identifies the left side as its characteristic function, including t=0. No first moment of this law has been used.

For $R>0$, logarithmic differentiation with the chain rule and the compact integral bridge give ${\int }_0^R\frac{x}{\pi (1+x^2)}dx=\frac{\log (1+R^2)}{2\pi }\mathrm{.}$ This tends to infinity: log is increasing by its positive derivative, and its inverse relation implies that log of an unbounded positive argument eventually exceeds every real level. Monotone convergence and density integration imply $\int \mathrm{\mid }x\mathrm{\mid }c(x)dx=+\mathrm{\infty }$, already from the positive half-line. The density is finite at x=0 and has total mass one; it is not a zero or point-mass law. AC is retained from density inversion and the compact integration bridge. The auxiliary Laplace density has a corner at zero, but all differentiations above were on individual half-lines and its use in inversion required continuity, not differentiability there.