# Solved – Entropy of Cauchy (Lorentz) Distribution

Entropy is defined as
$$H$$ = $$- int_chi p(x)$$ $$log$$ $$p(x)$$ $$dx$$

The Cauchy Distribution is defined as

$$f(x)$$ = $$frac{gamma}{pi}$$ $$frac{1}{gamma^2 + x^2}$$

I kindly ask to show the steps to calculate the Entropy of a Cauchy distribution, which is

$$log(4 pi gamma)$$

Reference: Cauchy distribution

Contents

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.

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