I was recently taught in a lecture on complex analysis based off of this paper that much of complex analysis can be rephrased in the language of Brownian Motion. The paper gives some simple proofs of standard results in complex analysis using this language, but I would be interested in seeing a proof of Cauchy's Integral formula or his residue theorem. In particular, I am curious about how one could compute an integral using the ideas of Brownian Motion.
Brownian Motion and Complex Analysis
brownian motioncomplex-analysis
Related Solutions
Other then as a fantastic tool to evaluate some difficult real integrals, complex integrals have many purposes.
Firstly, contour integrals are used in Laurent Series, generalizing real power series.
The argument principle can tell us the difference between the poles and roots of a function in the closed contour $C$:
$$\oint_{C} {f'(z) \over f(z)}\, dz=2\pi i (\text{Number of Roots}-\text{Number of Poles})$$
and this has been used to prove many important theorems, especially relating to the zeros of the Riemann zeta function.
Noting that the residue of $\pi \cot (\pi z)f(z)$ is $f(z)$ at all the integers. Using a square contour offset by the integers by $\frac{1}{2}$, we note the contour disappears as it gets large, and thus
$$\sum_{n=-\infty}^\infty f(n) = -\pi \sum \operatorname{Res}\, \cot (\pi z)f(z)$$
where the residues are at poles of $f$.
While I have only mentioned a few, basic uses, many, many others exist.
Solution using Fourier series
$$\begin{split} \int_{\mathbb R}\frac {|x|^p}{1+x^2}dx &= 2\int_0^{+\infty}\frac {x^p}{1+x^2}dx\\ &=2\left(\int_0^1\frac {x^p}{1+x^2}dx+\int_1^{+\infty}\frac {x^p}{1+x^2}dx\right)\\ &=\int_0^1\frac {u^{\frac p 2-\frac 1 2}}{1+u}du+\int_0^{1}\frac {y^{-\frac p2-\frac 1 2}}{1+y}dy \,\,\,\,\text{ with }u=x^2 \text{ and }y=\frac 1 {x^2} \end{split}$$ We thus need to compute, for $-1<\alpha<1$ $$\int_0^1\frac{v^{\alpha}}{1+v}dv =\int_0^1\sum_{n\in\mathbb N}(-1)^nv^{\alpha+n}dv=\sum_{n\in\mathbb N}(-1)^n\int_0^1v^{\alpha+n}dv=\sum_{n\in\mathbb N}\frac{(-1)^n}{\alpha+n+1}$$ In other words, $$\begin{split} \int_{\mathbb R}\frac {|x|^p}{1+x^2}dx &= \sum_{n\in\mathbb N}\frac{(-1)^n}{n+\frac p 2+\frac 1 2}+\sum_{n\in\mathbb N}\frac{(-1)^n}{n-\frac p 2+\frac 1 2}\\ &=\sum_{n\in\mathbb N}\frac{(-1)^n}{n-\frac p 2+\frac 1 2} + \sum_{n\in\mathbb N}\frac{(-1)^n}{-n+\frac p 2-\frac 1 2}\\ &= \sum_{n\in\mathbb N}\frac{(-1)^n}{n-\frac p 2+\frac 1 2} + \sum_{n\geq 1}\frac{(-1)^n}{-n-\frac p 2+\frac 1 2}\,\,\,\text{ (second index shifted by 1)}\\ &= \frac{1}{\frac 1 2 - \frac p 2} + (1-p)\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^2-\left(\frac 1 2-\frac p 2\right)^2} \end{split}$$ This classic sum can be computed via Fourier series, among other ways. For $s\notin \mathbb Z$, define $f_s$ as the $2\pi$-periodic function defined by $f_s(t)=\cos(st)$ for $|t|<\pi$. You can verify that its Fourier series expansion is $$\cos{st} = \frac{\sin{\pi s}}{\pi s} \left [1+2 s^2 \sum_{n\geq 1}\frac{(-1)^{n+1} \cos{n t}}{n^2-s^2} \right ]$$ Evaluating at $t=0$ gives $$\frac{\pi}{\sin(\pi s)}=\frac 1 s + 2s\sum_{n\geq 1}\frac{(-1)^{n+1} }{n^2-s^2}$$ Thus, plugging in $s=\frac 1 2 - \frac p 2$, $$\int_{\mathbb R}\frac {|x|^p}{1+x^2}dx=\frac {\pi}{\cos\left(\frac {\pi p} 2\right)}$$
Best Answer
I think that you are looking for the Kakutani Representation Theorem for harmonic functions: Let $B$ be a Brownian motion in $\mathbb{R}^n$ and let $U$ be a bounded open domain in $\mathbb{R}^n$. Let $x$ in $U$ and consider the Brownian motion $X(\cdot) = x + B(\cdot)$ starting at $x$. Let $\tau_x = \inf\{t\geq0\mid X(t) \notin U\}$ be the first hitting time of $\partial U$, which is finite a.s. Then for every continuous function $g:\partial U \to \mathbb{R}^n$ we have that
$$ u(x) = E\big{[}g(X(\tau_x))\big{]} $$ where $u$ is the unique solution to the Laplace equation in $U$ with boundary condition equal to $g$.
From this representation, note that the Mean Value Theorem for harmonic functions follows from the isotropy of the Brownian motion and the uniqueness of the solution to the Dirichlet problem. In the case $n=2$, you can use essentially the same argument to prove the Cauchy Integral Formula. Or, alternatively, you can deduce it from the Mean Value Theorem as in the answer here. (Once established for circles you can use the homotopy invariance of the integrals of holomorphic functions to extend the formula for arbitrary simple curves.)
Finally, regarding to how to compute an integral using Brownian motion, I recommend you take a look at the Feynman-Kac formula, which is one of the most beautiful results in mathematics that link two different worlds. It connects the worlds of Probability Theory and PDEs in a similar fashion to the Kakutani Representation Theorem above. Recall that the expected value of a random variable is in fact an integral so you can use the Kakutani Representation Theorem and the Feynman-Kac Formula to compute integrals.