Solved – Entropy of Cauchy (Lorentz) Distribution

As shown at How does entropy depend on location and scale?, the integral is easily reduced (via an appropriate change of variable) to the case $gamma=1$, for which

$$H = int_{-infty}^{infty} frac{log(1+x^2)}{1+x^2},dx.$$

Letting $x=tan(theta)$ implies $dx = sec^2(theta)dtheta$ whence, since $1+tan^2(theta) = 1/cos^2(theta)$,

$$H = -2int_{-pi/2}^{pi/2} log(cos(theta))dtheta = -4int_{0}^{pi/2} log(cos(theta))dtheta .$$

There is an elementary way to compute this integral. Write $I= int_{0}^{pi/2} log(cos(theta))dtheta$. Because $cos$ on this interval $[0, pi/2]$ is just the reflection of $sin$, it is also the case that $I= int_{0}^{pi/2} log(sin(theta))dtheta.$ Add the integrands:

$$logcos(theta) + logsin(theta) = log(cos(theta)sin(theta)) = log(sin(2theta)/2) = logsin(2theta) – log(2).$$

Therefore

$$2I = int_0^{pi/2} left(logsin(2theta) – log(2)right)dtheta =-frac{pi}{2} log(2) + int_0^{pi/2} logsin(2theta) dtheta.$$

Changing variables to $t=2theta$ in the integral shows that

$$int_0^{pi/2} logsin(2theta) dtheta = frac{1}{2}int_0^{pi} logsin(t) dt = frac{1}{2}left(int_0^{pi/2} + int_{pi/2}^piright)logsin(t)dt \= frac{1}{2}(I+I) = I$$

because $sin$ on the interval $[pi/2,pi]$ merely retraces the values it attained on the interval $[0,pi/2]$. Consequently $2I = -frac{pi}{2} log(2) + I,$ giving the solution $I = -frac{pi}{2} log(2)$. We conclude that

$$H = -4I = 2pilog(2).$$

An alternative approach factors $1+x^2 = (1 + ix)(1-ix)$ to re-express the integrand as

$$frac{log(1+x^2)}{1+x^2} = frac{1}{2}left(frac{i}{x-i} + frac{i}{x+i}right)log(1+ix) + frac{1}{2}left(frac{i}{x-i} + frac{i}{x+i}right)log(1-ix)$$

The integral of the first term on the right can be expressed as the limiting value as $Rtoinfty$ of a contour integral from $-R$ to $+R$ followed by tracing the lower semi-circle of radius $R$ back to $-R.$ For $Rgt 1$ the interior of the region bounded by this path clearly has a single pole only at $x=-i$ where the residue is

$$operatorname{Res}_{x=-i}left(left(frac{i}{x-i} + frac{i}{x+i}right)log(1+ix)right) = ileft.log(1 + ix)right|_{x=-i} = ilog(2),$$

whence (because this is a negatively oriented path) the Residue Theorem says

$$oint left(frac{1}{1+ix} + frac{1}{1-ix}right)log(1+ix) mathrm{d}x = -2pi i (ilog(2)) = 2pilog(2).$$

Because the integrand on the circle is $o(log(R)/R)$ which grows vanishingly small as $Rtoinfty,$ in the limit we obtain

$$int_{-infty}^infty frac{1}{2}left(frac{1}{1+ix} + frac{1}{1-ix}right)log(1+ix) mathrm{d}x = pilog(2).$$

The second term of the integrand is equal to the first (use the substitution $xto -x$), whence $H=2(pilog(2)) = 2pilog(2),$ just as before.