Question

By applying the residue theorem around a wedge-shaped contour of angle $2\pi /n$, with one side along the real axis, prove that the integral $${\int }_0^{\mathrm{\infty }}\frac{dx}{1+x^n}$$ where $n$ is real and $\ge 2$, has the value $(\pi /n)\mathrm{cosec}(\pi /n)$.

Short answer

The integral evaluates to $\frac{\pi }{n}\mathrm{cosec}(\pi /n)$.

Step-by-step solution

Define the integral and contour

Consider the integral ${\int }_0^{\mathrm{\infty }}\frac{dx}{1+x^n}$. Define a contour that starts from the origin, runs along the real axis to a point R, then an arc around the origin of angle $\frac{2\pi }{n}$ and radius R, and finally back to the origin along the real axis.

Apply the residue theorem

By the residue theorem, the integral over the full closed contour is equal to $2\pi i$ times the sum of residues of the integrand within the contour. Identify the poles inside the contour.

Identify the poles

The integrand $\frac{1}{1+z^n}$ has poles where $1+z^n=0$. Thus, the poles are at $z_k=e^{i(2k+1)\pi /n}$ for $k=0,1,...,n-1$.

Evaluate residue at the poles

The residue at each pole $z=z_k$ is given by $\text{Res}(\frac{1}{1+z^n},z=z_k)=\frac{1}{n{z}_k^{n-1}}$.

Compute the integral around the contour

The integral around the entire wedge-shaped contour equals the sum of integrals along each part. Show that the contributions from the arc and the line segments at infinity vanish as $R\to \mathrm{\infty }$.

Combine integrals and relate to given integral

Combine the integrals and relate them to the real axis integral. Use symmetry and limits to reduce the problem back to the original given integral.

Calculate the sum of residues and solve

Calculate the sum of residues at the poles and solve for the integral. The sum is $\sum _{k=0}^{n-1}\frac{1}{n{e}^{i(2k+1)\pi /n(n-1)}}$ which simplifies to $\frac{\pi }{n}\mathrm{cosec}(\pi /n)$.

Key concepts

Complex Analysis

Complex analysis is a branch of mathematics that studies functions of complex variables. These are numbers of the form $z=x+iy$, where $x$ and $y$ are real numbers, and $i$ is the imaginary unit. It focuses on integrals and derivatives of complex functions. This field has powerful theorems like the Cauchy Integral Theorem and the Residue Theorem.

Functions in complex analysis can have singularities, which are points where they become undefined. Understanding these singularities helps in evaluating contour integrals. These integrals go around paths in the complex plane Also, complex analysis uses techniques like contour integration for solving integrals with complex functions in a more straightforward manner than traditional methods.

Contour Integration

Contour integration involves integrating functions along paths, or 'contours,' in the complex plane. The chosen path can greatly simplify the evaluation of an integral. For our exercise, we used a wedge-shaped contour.

To set up a contour, we need:

• A path from one point to another, called the contour.
• The function which we will integrate.

In this example, our contour included:

• A line segment along the real axis.
• An arc of a circle centered at the origin.
• A radial line back to the origin.

By evaluating the integral along this path and using the symmetry and properties of our contour, we can simplify and solve the integral more easily.

Residues

Residues are a fundamental concept in complex analysis, playing a crucial role in evaluating complex integrals. At its core, the residue is the coefficient of $\frac{1}{z-a}$ in the Laurent series expansion of a function around a singularity. More informally, it captures the behavior of the function near singular points.

We use residues in conjunction with the Residue Theorem, which states: the integral of a function over a closed contour is equal to $2\pi i$ times the sum of residues inside that contour.

For our problem, the integrand $\frac{1}{1+z^n}$ had poles at specific points $z_k$. Calculating residues at each pole, we get $\text{Res}(\frac{1}{1+z^n},z=z_k)=\frac{1}{n{z}_k^{n-1}}$. Summing these residues helped in simplifying our integral.

Poles

Poles are specific kinds of singularities where a function's value tends to infinity. In the context of our problem, they are the values of $z$ that make the denominator zero in the function $\frac{1}{1+z^n}$. Poles are crucial for finding and evaluating residues.

Our poles are at points where the function has infinite behavior. Particularly, they occur at:

• $z_k=e^{i(2k+1)\frac{\tau }{n}}$
• Where $k=0,1,...,n-1$

Understanding where these poles occur and their nature allowed us to use the Residue Theorem effectively, ultimately leading to the solution of the integral.