Question

Show that \(\Gamma(z)=\int_{0}^{1}[\ln (1 / t)]^{z-1} d t\) with \(\operatorname{Re}(z)>0\).

Short answer

Through the substitution \(t = e^{-u}\) , we effectively show that \(\Gamma(z)=\int_{0}^{1}[\ln (1 / t)]^{z-1} dt\), where \(Re(z)>0\).

Step-by-step solution

Substitution

Begin by substituting \(t = e^{-u}\) into the gamma function. With this, \( dt = -e^{-u} du \). The integral limits change from \( (0, \infty)\) to \( (\infty, 0) \). The negative sign of \(dt\) makes the limits change their order.

Apply new substitution to Gamma function

Now the Gamma function becomes \( \Gamma(z) = \int_{0}^{\infty} [(-u)^{z-1} * e^{- u} * -du] \). We can write this integral in a form very similar to the given one: \( \Gamma(z) = \int_{0}^{1} [\ln(1 / t)]^{z-1} dt \)

Simplification and final form

We notice that the exponent of -u and -du cancel each other, and also -u to the power of z-1 becomes \(\ln(t)^{z-1}\), because the natural logarithm of \(e^{-u}\) just gives -u. Therefore, the gamma function simplifies to the given integral.

Key concepts

Complex Analysis

In the realm of complex analysis, we explore functions that operate within the complex number system. This area of mathematics is not only beautiful in its theoretical elegance but is also a cornerstone for numerous applications in engineering, physics, and other scientific disciplines. A key feature of complex analysis is dealing with integrals where the variables and bounds can take on complex values.

One such integral of significant interest is the Gamma function, defined for complex numbers with a positive real part. When confronted with the Gamma function, \(\Gamma(z)\), students often perform a substitution that transforms the integral into a more familiar territory. The substitution \(t = e^{-u}\) not only changes the variables in play but the integral's limits as well—shifting our perspective from the real number line to an exploration in the complex plane.

This practice of substitution is a shining example of the power of complex analysis: it allows us to reframe what might initially seem like an intractable problem into one that is solvable using our toolbox of integration techniques.

Gamma Function Properties

The Gamma function, \(\Gamma(z)\), which extends the concept of factorial to the complex numbers, boasts an array of fascinating properties, making it a frequent guest in mathematical problems and exercises. One crucial property, which the exercise demonstrates, is the integral representation of the Gamma function for complex numbers with positive real parts.

When you perform the change of variables \(t = e^{-u}\), you're effectively peeling back one layer of the function's properties, revealing a form that resonates with the function's definition as an integral. The fact that the exponent of \(u\) and the differential \(du\) cancel each other out is a testament to the function's internal consistency and the elegance of its properties. Students must understand these intrinsic qualities to manipulate and use the Gamma function effectively in various mathematical scenarios—from solving intricate integrals to exploring more profound aspects of functional analysis.

Mathematical Physics

In mathematical physics, the Gamma function is a vital tool, both for its direct implications and for the techniques it represents. The integration steps shown in the exercise mirror methods used in physics to tackle integrals encountered in quantum mechanics, thermodynamics, and statistical mechanics.

Role of the Gamma Function in Physics

The function's ability to handle factorial-like operations for complex numbers is especially pertinent when dealing with probability distributions that arise in various physical systems. For example, in quantum mechanics, the normalization conditions of wave functions may involve integrals that can be resolved using properties of the Gamma function.

Technique Transfer

Furthermore, the substitution technique demonstrated ( \(t = e^{-u}\) ) is akin to methods employed in transforming differentials and variables. This skill is essential for physicists as they often face the challenge of simplifying complex integrals to reveal physically interpretable forms. Understanding the Gamma function and its properties is thus more than a mathematical exercise—it’s an illustration of the essential methodologies physicists rely on to decode the language of the universe.