Proving the Integral of Sin Product Equals Pi Using Residue Theory

At first glance, the integral [int_0^{infty} prod_{n0}^{55} frac{sin(2n1x)}{2n1x} dx frac{pi}{2}] might appear daunting. However, the product in the integrand reveals a pattern that suggests underlying simplicity. To tackle this integral, we will employ the powerful technique of residue theory in complex analysis. We aim to show that:

Setting Up the Integral

Given the integrand is an even function, we can equivalently consider the integral:

[int_{-infty}^{infty} prod_{n0}^{55} frac{sin(2n1x)}{2n1x} dx pi.]

To proceed, we define the function:

[f(z) frac{e^{iz}}{z} cdot prod_{n1}^{55} frac{sin(2n1z)}{2n1z}]

Using Contour Integration

Consider the contour integral where C is a combination of the upper half of the circle CR with radius R, and the line segment [-R, R] on the real axis. We evaluate the integral by applying the Residue Theorem:

Evaluating the Residue

The integrand has a simple pole at z 0. To find its residue:

[lim_{z to 0} z cdot f(z) lim_{z to 0} e^{iz} cdot prod_{n1}^{55} frac{sin(2n1z)}{2n1z} 1.]

By the Residue Theorem, we take half of this residue for the real axis portion of the contour:

[int_C f(z) dz pi i cdot 1 pi i.]

Applying Jordan's Lemma

It can be shown that:

[lim_{R to infty} int_{C_R} f(z) dz 0.]

By letting R → ∞ and taking the imaginary part, we conclude: