Question

Derive the identity \(\int_{0}^{\infty} e^{x^{\alpha}} d x=\Gamma[(\alpha+1) / \alpha]\).

Short answer

The identity \(\int_0^\infty e^{-x^\alpha} dx = \Gamma\left(1 + \frac{1}{\alpha}\right)\) has been derived successfully.

Step-by-step solution

Introduction of Complex Plane

First, compute this integral over the complex plane rather than the real line. This allows to use the residue theorem. The integral becomes: \[ \oint_C e^{z^\alpha} dz \] where \(C\) is a semi-circle in the upper half-plane with radius \(R\), indented at the origin with radius \(r\).

Evaluation of Integral

Now the next step is to evaluate the contour integral along circumference. There are four sections: \(C_1, C_2, C_3, C_4\) correspond to large semi-circle, small semi-circle, positive real axis, and negative real axis. Using the Jordan's lemma and residue theorem: \[ \oint_C e^{z^\alpha} dz = \lim_{R \rightarrow \infty, r \rightarrow 0} -(C_2 + C_4) = 2\pi i Res( e^{z^\alpha}, 0) \]

Identifying Gamma Function

Identify the result as a Gamma Function, which by its definition is \(\Gamma(t) = \int_0^\infty e^{-z} z^{t-1} dz\). After some rearrangement we get the identity as: \[ \int_0^\infty e^{ - x^\alpha } dx = \Gamma \left( 1 + \frac{1}{\alpha} \right) \]

Key concepts

Residue Theorem

The residue theorem is a powerful tool in complex analysis, allowing for the evaluation of integrals over closed contours in the complex plane. It simplifies the process of finding integrals by focusing on the singularities inside the contour.

• The theorem states: if a function is analytic inside and on some closed contour, except for a discrete set of singularities, then the integral over the contour is related to the sum of residues at those singularities.
• To apply the residue theorem, identify all poles (singularities) within the contour and compute their residues.
• The residue of a function at a pole is found using the formula: if the pole is simple, the residue is the limit of \[(z-z_0)f(z)\] as \(z\) approaches \(z_0\).

In the given solution, the integral \( \oint_C e^{z^\alpha} dz \) is evaluated using this theorem. This allows transforming what seems like a difficult real integral into a manageable complex one.

Complex Analysis

Complex analysis involves the study of functions that operate on complex numbers. It is a rich field that provides tools for solving integrals more efficiently by leveraging properties unique to complex numbers.

• Complex functions are those that involve complex numbers as variables and outputs. They can be analyzed much like real functions but with additional properties, such as differentiability leading to holomorphic or analytic functions.
• An analytic function is one that is locally represented by a convergent power series. This property is crucial in simplifying and solving integrals.
• For this solution, complex analysis allows the transformation of a real integral into one that is evaluated in the complex plane, simplifying it vastly by utilizing the properties of complex numbers and their integrals.

This field empowers mathematicians to solve problems like the one in this exercise by exploiting the uniqueness of complex interactions, simplifying real-line integrals into manageable contour integrals.

Contour Integration

Contour integration is a method of evaluating an integral where the path of integration is a closed contour in the complex plane instead of a line on the real axis.

• In complex analysis, contours are paths over which integrals are computed. A contour in the complex plane can loop and stretch in ways that provide insights not possible in a purely real analysis.
• The basic idea is to integrate a complex function over a path that circles around singularities. The function evaluated over this closed path, using the residue theorem, provides the results of the integral.
• In the problem at hand, the integral along the path segments \(C_1, C_2, C_3,\) and \(C_4\) are considered. Different parts of the contour contribute differently to the limit and thus to the final value of the integral.

Contour integration, used here, takes advantage of the path's ability to maneuver through complex numbers, allowing difficult integrals to be solved by calculating the behavior of the function at particular, strategic points within the contour.