Question

Prove, for every \(p, q>-1, \quad \int_{0}^{1} x^{p}\left(\ln \frac{1}{x}\right)^{q} d x=\frac{\Gamma(q+1)}{(p+1)^{q+1}}\)

Short answer

The equation is proven by converting the integral into a Gamma function form.

Step-by-step solution

Recognize the Gamma Function

The Gamma Function is denoted as \(\Gamma(z)\) and is defined as \(\Gamma(z) = \int_0^\infty t^{z-1} e^{-t} dt\) for \(z > 0\). Recognize that transforming the given integral into a form that uses the Gamma function will be beneficial.

Substitute Variables

Use the substitution \(x = e^{-t}\). Then \(dx = -e^{-t} dt\), or equivalently, \(dt = -\frac{dx}{x}\). When \(x = 0\), \(t = +\infty\); and when \(x = 1\), \(t = 0\). Hence, the integral becomes \(\int_\infty^0 e^{-pt} t^q (-dt)\).

Rearrange Integral Limits

Change the limits of integration to match the standard form: \[ \int_0^\infty e^{-pt} t^q dt. \] Notice this looks similar to the Gamma function with a scaling factor.

Modify the Integrand

Modify the integrand by introducing a substitution to simplify it to match the Gamma function. Let \(u = pt\), then \(t = \frac{u}{p}\) and \(dt = \frac{1}{p} du\).

Final Formulation

The integral in terms of \(u\) becomes: \[ \int_0^\infty \left(\frac{u}{p}\right)^q e^{-u} \frac{1}{p} du = \frac{1}{p^{q+1}} \int_0^\infty u^q e^{-u} du. \] This can be recognized as \(\frac{1}{p^{q+1}} \Gamma(q+1)\).

Conclusion

From the previous step, we observe \(\Gamma(q+1)\) multiplied by \(\frac{1}{p^{q+1}}\) matches the form of the Gamma function integral. Thus the evaluated integral is \(\frac{\Gamma(q+1)}{(p+1)^{q+1}}\), confirming the original statement.

Key concepts

Gamma Function

The Gamma Function is an extension of the factorial function to real and complex numbers. It is denoted as \( \Gamma(z) \) and is defined for positive real numbers, \( z > 0 \), by the integral: \[ \Gamma(z) = \int_0^\infty t^{z-1} e^{-t} dt. \] This function is fundamental in the fields of mathematics and physics because it frequently appears in many areas including probability, statistics, and complex analysis.

• The Gamma function satisfies \( \Gamma(n) = (n-1)! \) for a positive integer \( n \).
• It shows us how we can generalize the notion of factorial beyond integers.

In the given exercise, the goal is to express the integral in a form that aligns with the Gamma Function to facilitate the proof. Understanding and utilizing the properties of the Gamma function allow us to bridge from known mathematical concepts (like factorials) to broader, more abstract concepts in integration and calculus.

Substitution Method

The substitution method in calculus is a technique used to simplify the process of integrating complex expressions. The goal is to transform the integral into a standard form that is easier to evaluate. In this exercise, the substitution \( x = e^{-t} \) was used. This leads to:

• The derivative: \( dx = -e^{-t} dt \) or \( dt = -\frac{dx}{x} \).
• Altering limits of integration: when \( x = 0 \), \( t = +\infty \) and when \( x = 1 \), \( t = 0 \).

After substitution, the integral is expressed in a new variable, \( t \), making it easier to manipulate. This change helps to simplify the integration limits and transform the integral into a recognizable form that aligns with known functions and integrals, such as the Gamma Function.

Definite Integrals

Definite integrals are used to compute the total accumulation of quantities, like area under a curve. They are represented as \( \int_a^b f(x) dx \), where \( a \) and \( b \) are the limits of integration. For this integral, the strategy was to adjust the integral into a known form. Initially, the problem’s integral was rewritten through substition, with the limits adjusted:

• Initial form: \( \int_{0}^{1} x^{p}\left(\ln \frac{1}{x}\right)^{q} dx \).
• Modified form: \( \int_0^\infty e^{-pt} t^q dt \) after substitution.

The rewritten integral allowed for further manipulations with another substitution to ultimately identify a clear connection with the Gamma function. Definite integrals can sometimes prove challenging, but by transforming the offered expression, they can transform into recognizable and simpler forms.

Mathematical Proofs

Mathematical proofs are logical arguments that confirm the truth of a mathematical statement. A good proof not only confirms a statement but also explains why it is true. In this exercise, the proof involves:

• Introducing a substitution, thereby altering the form and limits of an integral.
• Recognizing forms related to common mathematical functions, like the Gamma Function.
• Utilizing known formulas and properties (e.g., properties of exponential functions and integrals).

Through step-by-step transformations, the original integral is manipulated to eventually show it is equivalent to an expression involving the Gamma Function, which was the target to prove. Mathematical proofs hinge on logical progression and transformations that often require creative but precise steps to confirm the final assertion.