Question

By considering the integral of $${\left(\frac{\sin \alpha z}{\alpha z}\right)}^2\frac{\pi }{\sin \pi z},\alpha <\frac{\pi }{2}$$ around a circle of large radius, prove that $$\sum _{m=1}^{\mathrm{\infty }}(-1{)}^{m-1}\frac{{\sin }^{2}m\alpha }{(m\alpha {)}^2}=\frac{1}{2}$$

Short answer

By summing residues using the residue theorem, it shows that $\sum _{m=1}^{\mathrm{\infty }}(-1{)}^{m-1}\frac{{\sin }^{2}m\alpha }{(m\alpha {)}^2}=\frac{1}{2}$.

Step-by-step solution

Define the Integral

Consider the integral $$I={\oint }_C\frac{(\frac{\sin \alpha z}{\alpha z}{)}^2\pi }{\sin \pi z}dz$$ where $C$ is a circle of large radius centered at the origin.

Analyze the Poles and Residues

Examine the integrand for its poles. They occur at the points where the denominator $\sin \pi z=0$, i.e., at $z=n$ for any integer $n$. Determine the residues at these poles.

Residue Calculation

To find the residue at a pole $z=n$, use the Laurent series expansion. For a simple pole at $z=m$: $$\text{Residue}=\underset{z\to m}{lim}(z-m)\frac{(\frac{\sin \alpha z}{\alpha z}{)}^2\pi }{\sin \pi z}$$

Sum of Residues

Since the residues for nonzero integers are what contribute to the integral, sum all residues at $me0$. The series representation for these residues leads to: $${\text{Residue}}_m=(-1{)}^{m-1}\frac{{\sin }^{2}m\alpha }{m^2{\alpha }^2}$$

Apply Residue Theorem

Apply the residue theorem: $$I=2\pi i\sum _{m=1}^{\mathrm{\infty }}(-1{)}^{m-1}\frac{{\sin }^{2}m\alpha }{(m\alpha {)}^2}$$

Evaluate the Integral

Consider the contribution of the integral on the large circle. As the radius grows, given $\alpha <\frac{\pi }{2}$, the integral over this contour tends to zero.

Solve the Equation

The full integral equates to zero, therefore: $$2\pi i\sum _{m=1}^{\mathrm{\infty }}(-1{)}^{m-1}\frac{{\sin }^{2}m\alpha }{(m\alpha {)}^2}=0$$ Dividing both sides by $2\pi i$ results in: $$\sum _{m=1}^{\mathrm{\infty }}(-1{)}^{m-1}\frac{{\sin }^{2}m\alpha }{(m\alpha {)}^2}=\frac{1}{2}$$

Key concepts

Residue Theorem

The Residue Theorem is a powerful tool in complex analysis. It allows us to evaluate complex integrals by examining the residues of poles inside a contour. This theorem states that for a meromorphic function, the integral around a closed contour is equal to $2\pi i$ times the sum of the residues inside that contour. The formula is:

$${\oint }_{C}f(z)dz=2\pi i\sum _{\text{residues}}\text{Res}(f,z_k)$$

In our exercise, we used the Residue Theorem to calculate the integral by summing the residues at the poles of the function. This method simplifies solving integrals that would otherwise be very complicated.

Laurent Series

The Laurent Series is an expansion of a complex function that includes terms with negative powers of the variable. Unlike Taylor series, which are always centered around a point free of singularities, Laurent series can represent functions with poles. It takes the form:

$$f(z)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_n(z-z_0{)}^n$$

In the exercise, we used the Laurent series to find the residue at a pole. The residue is simply the coefficient of the $\frac{1}{z}$ term in the Laurent series expansion. To identify this term, we applied the Laurent series to the function's components to isolate the residues.

Poles and Residues

Poles are specific types of singularities where a complex function approaches infinity. The order of a pole is determined by how fast the function grows near the pole. A first-order pole (or simple pole) satisfies:

$$\underset{z\to z_0}{lim}(z-z_0)f(z)=\text{finite}$$

To find residues at these poles, one method is:

• Identify where the function has poles.
• For each pole, find the residue, particularly important for first-order poles.

An example from our exercise was finding residues at poles of $$z=n\text{where}n\in \mathbb{Z}.$$ Calculating these residues helps sum them up in the integral, which is crucial for using the Residue Theorem.

Contour Integration

Contour integration involves integrating a complex function along a specified path (or contour) in the complex plane. This technique is valuable for dealing with integrals in complex analysis, particularly when the function has singularities. The basic steps for contour integration are:

• Define the integral over a closed contour.
• Examine the integrand for singularities (poles).
• Apply relevant theorems (like Cauchy’s Integral Formula or Residue Theorem).
• Calculate residues and sum them up.

For the exercise, we considered an integral around a large circle. By analyzing the residues of our function within this contour and accounting for the contours contribution at the infinite boundary, we used contour integration to solve the integral effectively.