Evaluating the Improper Integral of (sin x)/x Using Complex Analysis and Fubini’s Theorem

Mathematically evaluating the integral of the function (sin x)/x (also known as the sinc function) is a fascinating problem with no elementary antiderivative. In this article, we will explore two advanced methods: Fubini’s theorem and complex analysis. Both methods offer a unique way to understand and solve this intriguing integral problem.

Introduction to Fubini's Theorem

Let's begin by discussing the technique involving Fubini’s theorem. Fubini's theorem provides a method to interchange the order of integration in double integrals under certain conditions. This theorem is particularly useful in dealing with functions that can be integrated over a region in multiple ways.

Step-by-Step Solution Using Fubini's Theorem

Consider the integral: $$int_{0}^{infty} frac{sin x}{x} , dx.$$

First, we note that the integrand is even, meaning (frac{sin x}{x}) is symmetric about the y-axis. Hence, we can write:

$$int_{0}^{infty} frac{sin x}{x} , dx frac{1}{2} int_{-infty}^{infty} frac{sin x}{x} , dx.$$

Now, we use a key property of the integrand:

$$frac{1}{x} int_{0}^{infty} e^{-xt} , dt.$$

This allows us to rewrite our integral as:

$$int_{0}^{infty} frac{sin x}{x} , dx frac{1}{2} int_{0}^{infty} left(int_{0}^{infty} e^{-xt} sin x , dtright) dx.$$

To interchange the order of integration, we use Fubini's theorem. This gives us:

$$int_{0}^{infty} frac{sin x}{x} , dx frac{1}{2} int_{0}^{infty} left(int_{0}^{infty} sin x , e^{-xt} , dxright) dt.$$

Let's denote the inner integral as (I_1):

$$I_1 int_{0}^{infty} sin x , e^{-xt} , dx.$$

Integrating (I_1) by parts, we get:

$$I_1 frac{1}{t} int_{0}^{infty} e^{-xt} cos x , dx.$$

Integrating by parts again, we obtain:

$$int_{0}^{infty} e^{-xt} cos x , dx frac{1}{t^2} - frac{1}{t^2} int_{0}^{infty} e^{-xt} sin x , dx frac{1}{t^2} - frac{1}{t^2} I_1.$$

Solving for (I_1), we find:

$$I_1 frac{1}{1 t^2}.$$

Therefore:

$$int_{0}^{infty} frac{sin x}{x} , dx frac{1}{2} int_{0}^{infty} frac{1}{1 t^2} , dt frac{1}{2} left. arctan(t) right|_{0}^{infty} frac{1}{2} left(frac{pi}{2} - 0right) frac{pi}{4}.$$

Since our original integral was (frac{1}{2} int_{-infty}^{infty} frac{sin x}{x} , dx), multiplying by 2, we get:

$$int_{-infty}^{infty} frac{sin x}{x} , dx pi.$$

Approach Using Complex Analysis

Next, we will use complex analysis to evaluate the same integral. We will consider the complex function (f(z) frac{exp(iz)}{z}) and use Cauchy's integral theorem and the residue theorem.

Step-by-Step Solution Using Complex Analysis

Consider the integral:

$$int_{-infty}^{infty} frac{exp(iz)}{z} , dz.$$

This function is holomorphic except at the point (z 0), which is a simple pole. We integrate along a closed contour made up of four segments: (gamma_1), (gamma_2), (gamma_3), and (gamma_4).

By Cauchy's integral theorem, the integral over the closed contour is zero:

$$oint_{gamma} f(z) , dz 0.$$

Breaking it down, we get:

$$int_{gamma_1} f(z) , dz int_{gamma_2} f(z) , dz int_{gamma_3} f(z) , dz int_{gamma_4} f(z) , dz 0.$$

As (R to infty), the integral along (gamma_2) (the arc) goes to zero:

$$lim_{R to infty} int_{gamma_2} f(z) , dz lim_{R to infty} int_{pi}^{2pi} frac{exp(iR e^{itheta})}{R e^{itheta}} iR e^{itheta} , dtheta int_{pi}^{2pi} i exp(iR e^{itheta}) , dtheta 0.$$

Next, we calculate the integral along (gamma_4):

$$lim_{epsilon to 0} int_{gamma_4} f(z) , dz lim_{epsilon to 0} int_{pi}^{2pi} frac{exp(i epsilon e^{-i t})}{epsilon e^{-i t}} (-i epsilon e^{-i t}) , dt -i pi.$$

Combining the integrals along (gamma_3) and (gamma_1) (the real axis), we get:

$$int_{-infty}^{infty} f(z) , dz - i pi 0 Leftrightarrow int_{-infty}^{infty} frac{exp(iz)}{z} , dz i pi.$$

Looking only at the imaginary part, we obtain: