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Example 2 – Cauchy distribution

The integral ∫ − ∞ ∞ e i t x x 2 + 1 d x {\displaystyle \int _{-\infty }^{\infty }{\frac {e^{itx}}{x^{2}+1}}\,dx}

(which arises in probability theory as a scalar multiple of the characteristic function of the Cauchy distribution) resists the techniques of elementary calculus. We will evaluate it by expressing it as a limit of contour integrals along the contour C that goes along the real line from − a to a and then counterclockwise along a semicircle centered at 0 from a to − a. Take a to be greater than 1, so that the imaginary unit i is enclosed within the curve. The contour integral is ∫ C e i t z z 2 + 1 d z. {\displaystyle \int _{C}{\frac {e^{itz}}{z^{2}+1}}\,dz.}

Since e is an entire function (having no singularities at any point in the complex plane), this function has singularities only where the denominator z + 1 is zero. Since z + 1 = (z + i)(z − i), that happens only where z = i or z = − i. Only one of those points is in the region bounded by this contour. The residue of f (z) at z = i is lim z → i (z − i) f (z) = lim z → i (z − i) e i t z z 2 + 1 = lim z → i (z − i) e i t z (z − i) (z + i) = lim z → i e i t z z + i = e − t 2 i. {\displaystyle \lim _{z\to i}(z-i)f(z)=\lim _{z\to i}(z-i){\frac {e^{itz}}{z^{2}+1}}=\lim _{z\to i}(z-i){\frac {e^{itz}}{(z-i)(z+i)}}=\lim _{z\to i}{\frac {e^{itz}}{z+i}}={\frac {e^{-t}}{2i}}.}

According to the residue theorem, then, we have ∫ C f (z) d z = 2 π i Res z = i f (z) = 2 π i e − t 2 i = π e − t. {\displaystyle \int _{C}f(z)\,dz=2\pi i\operatorname {Res} _{z=i}f(z)=2\pi i{\frac {e^{-t}}{2i}}=\pi e^{-t}.}

The contour C may be split into a "straight" part and a curved arc, so that ∫ straight + ∫ arc = π e − t, {\displaystyle \int _{\text{straight}}+\int _{\text{arc}}=\pi e^{-t},} and thus ∫ − a a = π e − t − ∫ arc. {\displaystyle \int _{-a}^{a}=\pi e^{-t}-\int _{\text{arc}}.}

According to Jordan's lemma, if t > 0 then ∫ arc e i t z z 2 + 1 d z → 0 as a → ∞. {\displaystyle \int _{\text{arc}}{\frac {e^{itz}}{z^{2}+1}}\,dz\rightarrow 0{\mbox{ as }}a\rightarrow \infty.}

Therefore, if t > 0 then ∫ − ∞ ∞ e i t x x 2 + 1 d x = π e − t. {\displaystyle \int _{-\infty }^{\infty }{\frac {e^{itx}}{x^{2}+1}}\,dx=\pi e^{-t}.}

A similar argument with an arc that winds around − i rather than i shows that if t < 0 then ∫ − ∞ ∞ e i t x x 2 + 1 d x = π e t, {\displaystyle \int _{-\infty }^{\infty }{\frac {e^{itx}}{x^{2}+1}}\,dx=\pi e^{t},} and finally we have this: ∫ − ∞ ∞ e i t x x 2 + 1 d x = π e − | t |. {\displaystyle \int _{-\infty }^{\infty }{\frac {e^{itx}}{x^{2}+1}}\,dx=\pi e^{-|t|}.}

(If t = 0 then the integral yields immediately to real-valued calculus methods and its value is π.)