Question

Express the following as Gamma functions. Namely, noting the form \(\Gamma(x+1)=\int_{0}^{\infty} t^{x} e^{-t} d t\) and using an appropriate substitution, each expression can be written in terms of a Gamma function. a. \(\int_{0}^{\infty} x^{2 / 3} e^{-x} d x\) b. \(\int_{0}^{\infty} x^{5} e^{-x^{2}} d x\) c. \(\int_{0}^{1}\left[\ln \left(\frac{1}{x}\right)\right]^{n} d x\)

Short answer

The integrals as Gamma functions are: (a) \(3 \Gamma(\frac{11}{3})\), (b) \(\frac{1}{2} \Gamma(3.5)\), (c) \(\Gamma(n+1)\).

Step-by-step solution

Solution for part (a)

The integral is \(\int_{0}^{\infty} x^{2 / 3} e^{-x} d x\). For substitution, let \(x = t^3\). Then, \(dx = 3t^2 dt\) and the integral becomes \(\int_{0}^{\infty} t^{2} e^{-t^3} 3t^2 dt = 3 \int_{0}^{\infty} t^{8/3} e^{-t^3} dt\). But this integral is not in the form of Gamma function, so we need another substitution. Let's call \(x = u^3\). Then \(dx = 3u^2 du\) and the integral becomes \(\int_{0}^{\infty} u^{8/3} e^{-u} du\). Comparing this with the form of the Gamma function, we can see that this is equivalent to \(3 \Gamma(\frac{11}{3})\).

Solution for part (b)

The integral is \(\int_{0}^{\infty} x^{5} e^{-x^{2}} d x\). For substitution, let \(x^2 = t\). Then \(dx = \frac{1}{2}t^{-1/2} dt\) and the integral becomes \(\int_{0}^{\infty} t^{2.5} e^{-t} dt / 2\). Comparing this with the form of the Gamma function, this integral is equivalent to \(\frac{1}{2} \Gamma(3.5)\).

Solution for part (c)

The integral \(\int_{0}^{1}\left[\ln \left(\frac{1}{x}\right)\right]^{n} d x\) and substitute \(u=\ln(1/x)\). So, \(-x du=dx\) which transforms the integral to \(\int_{0}^{\infty} u^{n} e^{-u} du\). After comparing this with the form of the Gamma function, we can see this integral is equivalent to \(\Gamma(n+1)\).

Key concepts

Mathematical Integration

Mathematical integration is a fundamental concept in calculus that deals with the process of finding the integral of a function. This process can be viewed as the opposite of differentiation, and it involves finding the total value that a function accumulates over a particular interval.

For example, in the Gamma function exercise, integration is used to determine the area under the curve of the function, which in this case involves an exponential decay term, represented by the exponential function. Integration helps in simplifying complex functions into more understandable forms and is especially handy when dealing with infinite series or limits, as in the case of the Gamma function which involves an integral from zero to infinity.

A common integration technique is integration by substitution, which is used when a function is too complex to integrate in its current form. By substituting parts of the function with a new variable, the integral can often be transformed into a simpler form that is more easily solved. This is exemplified in the solutions provided for the exercise, where substitutions were made to write the integrals in terms of the Gamma function, a process which we'll cover more in the 'Integration by Substitution' section.

Exponential Functions

Exponential functions are mathematical expressions where a constant base is raised to a variable exponent. These functions are crucial in modeling scenarios where growth or decay accelerates rapidly, such as in population dynamics or radioactive decay.

In the context of the Gamma function exercise, an exponential function takes the form of the term e^-x or e^-t, where e is the base of the natural logarithm and the exponent is a variable representing a negative decay rate. This function decreases as x increases, which is characteristic of a decay process.

The interaction between exponential functions and powers of variables are central to the structure of the Gamma function, impacting how the integral is approached and solved. The exponential function in the Gamma function is essential as it makes the improper integral (an integral with one or more infinite limits) converge instead of diverging to infinity.

Integration by Substitution

Integration by substitution is a technique used to evaluate integrals that might not be immediately integrable in their original form. It involves a change of variables to transform the integral into a form that is more straightforward to integrate.

For instance, consider part (a) of the exercise. The initial substitution x = t³ seemed promising but didn’t directly give a Gamma function form. A subsequent substitution was needed to meet the conditions of the Gamma function’s definition.

This method is sometimes compared to the chain rule in differentiation, as it deals with composite functions (functions within functions). In the exercise, you’ll notice for part (b) that x² = t was used because the integral contained x⁵e^-x2, and taking the square root of t simplifies the power of x. Similarly, for part (c), the substitution u = ln(1/x) turns the integral into a form that matches the Gamma function template provided in the problem statement.

Such strategic substitutions shape the integral into a form where the Gamma function's formula can be directly applied, enabling the solution to be expressed neatly in terms of known Gamma function values.