Residue theorem
An integral along the real axis

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

arises in probability theory when calculating the characteristic function of the Cauchy distribution. It resists the techniques of elementary calculus but can be evaluated by expressing it as a limit of contour integrals.

Suppose t > 0 and define 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. Now consider the contour integral ∫ C f (z) d z = ∫ C e i t z z 2 + 1 d z. {\displaystyle \int _{C}{f(z)}\,dz=\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. Because f (z) is e i t z z 2 + 1 = e i t z 2 i (1 z − i − 1 z + i) = e i t z 2 i (z − i) − e i t z 2 i (z + i), {\displaystyle {\begin{aligned}{\frac {e^{itz}}{z^{2}+1}}&={\frac {e^{itz}}{2i}}\left({\frac {1}{z-i}}-{\frac {1}{z+i}}\right)\\&={\frac {e^{itz}}{2i(z-i)}}-{\frac {e^{itz}}{2i(z+i)}},\end{aligned}}} the residue of f (z) at z = i is Res z = i f (z) = e − t 2 i. {\displaystyle \operatorname {Res} _{z=i}f(z)={\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\cdot \operatorname {Res} \limits _{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 ∫ s t r a i g h t f (z) d z + ∫ a r c f (z) d z = π e − t {\displaystyle \int _{\mathrm {straight} }f(z)\,dz+\int _{\mathrm {arc} }f(z)\,dz=\pi e^{-t}} and thus ∫ − a a f (z) d z = π e − t − ∫ a r c f (z) d z. {\displaystyle \int _{-a}^{a}f(z)\,dz=\pi e^{-t}-\int _{\mathrm {arc} }f(z)\,dz.}

Using some estimations, we have | ∫ a r c e i t z z 2 + 1 d z | ≤ π a ⋅ sup arc | e i t z z 2 + 1 | ≤ π a ⋅ sup arc 1 | z 2 + 1 | ≤ π a a 2 − 1, {\displaystyle \left|\int _{\mathrm {arc} }{\frac {e^{itz}}{z^{2}+1}}\,dz\right|\leq \pi a\cdot \sup _{\text{arc}}\left|{\frac {e^{itz}}{z^{2}+1}}\right|\leq \pi a\cdot \sup _{\text{arc}}{\frac {1}{|z^{2}+1|}}\leq {\frac {\pi a}{a^{2}-1}},} and lim a → ∞ π a a 2 − 1 = 0. {\displaystyle \lim _{a\to \infty }{\frac {\pi a}{a^{2}-1}}=0.}

The estimate on the numerator follows since t > 0, and for complex numbers z along the arc (which lies in the upper half-plane), the argument φ of z lies between 0 and π. So, | e i t z | = | e i t | z | (cos φ + i sin φ) | = | e − t | z | sin φ + i t | z | cos φ | = e − t | z | sin φ ≤ 1. {\displaystyle \left|e^{itz}\right|=\left|e^{it|z|(\cos \varphi +i\sin \varphi)}\right|=\left|e^{-t|z|\sin \varphi +it|z|\cos \varphi }\right|=e^{-t|z|\sin \varphi }\leq 1.}

Therefore, ∫ − ∞ ∞ e i t z z 2 + 1 d z = π e − t. {\displaystyle \int _{-\infty }^{\infty }{\frac {e^{itz}}{z^{2}+1}}\,dz=\pi e^{-t}.}

If t < 0 then a similar argument with an arc C ′ that winds around − i rather than i shows that

∫ − ∞ ∞ e i t z z 2 + 1 d z = π e t, {\displaystyle \int _{-\infty }^{\infty }{\frac {e^{itz}}{z^{2}+1}}\,dz=\pi e^{t},}

and finally we have ∫ − ∞ ∞ e i t z z 2 + 1 d z = π e − | t |. {\displaystyle \int _{-\infty }^{\infty }{\frac {e^{itz}}{z^{2}+1}}\,dz=\pi e^{-\left|t\right|}.}

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