Question

Use integration by parts to show \(\Gamma(x+1)=x \Gamma(x)\)

Short answer

The identity \(\Gamma(x+1)=x \Gamma(x)\) is indeed correct. We have proved it using integration by parts.

Step-by-step solution

Express \(\Gamma(x+1)\) through the integral

Begin with writing the Gamma function for \((x+1)\): \(\Gamma(x+1)=\int_0^{\infty} t^{x} e^{-t} dt\).

Split the integral with respect to differentiation product rule

Next, it is commonly known that the integral of a product can be expressed via integration by parts, where \(u = t^{x}\) and \(dv = e^{-t} dt\). Differentiate \(u\) and integrate \(dv\) to get necessary components of the formula. We obtain: \(du = x t^{x-1} dt\) and \(v = -e^{-t}\).

Apply the integration by parts

Following the formula for integration by parts \(\int u dv = uv - \int v du\), replace \(u\), \(v\), \(du\) in this formula. We get \(\int_0^{\infty} t^{x} e^{-t} dt = -t^{x} e^{-t} |_{0}^{\infty} - \int_0^{\infty} -x t^{x-1} e^{-t} dt\). The expression in the absolute values vanishes for both infinity and zero (lim_{t->0} -t^{x} e^{-t} = 0 and lim_{t->infinity} -t^{x} e^{-t} = 0), this leaves us with \(\Gamma(x+1) = x \int_0^{\infty} t^{x-1} e^{-t} dt\).

Conclude the proof

The right hand side of the above equation is exactly \(x \Gamma(x)\) as asked in the exercise.

Key concepts

Integration by Parts

The method of integration by parts is a powerful tool in calculus, especially for solving integrals of products of functions. The technique is based on the product rule for differentiation and is generally used when the integral cannot be computed directly. When facing an integral of the form \(\int u dv\), integration by parts allows us to transform it into a presumably simpler form: \( uv - \int v du\).

The key step is to judiciously choose \(u\) and \(dv\) so that differentiating \(u\) reduces its order or complexity and integrating \(dv\) remains manageable. When applying this strategy to the Gamma function as seen in the exercise, the selection of \(u = t^{x}\) and \(dv = e^{-t} dt\) is strategic and clever as it simplifies the integral on the right-hand side into a form directly related to the Gamma function itself, showcasing integration by parts as the solution mechanism.

Mathematical Induction

Mathematical induction is a technique used to prove a statement, formula, or theorem which is asserted to be true for all natural numbers. It consists of two steps: the base case and the induction step. In the base case, the statement is proven for the initial value, often \( n=0 \) or \( n=1\). The induction step, however, requires one to assume that the statement holds for some arbitrary natural number \( k\) and then prove that the statement also holds for \( k+1\).

This form of reasoning is incredibly powerful for proving propositions related to sequences, series, and, as in our exercise, properties of special functions like the Gamma function. While mathematical induction wasn't directly used in this exercise, understanding this technique is crucial as it often complements arguments where a generalization from one case to the next is required.

Special Functions

Special functions, such as the Gamma function, play a vital role in various areas of mathematics, physics, and engineering. They arise in solutions to differential equations, complex analysis, and in various series expansions. The Gamma function, denoted as \( \Gamma(x) \), extends the concept of factorial to real and complex numbers. It is defined as \( \Gamma(x) = \int_0^{\infty} t^{x-1} e^{-t} dt \) for \( x > 0 \).

Special functions often possess intriguing properties, such as recurrence relations, which allow for simplification and computation in complex case scenarios. In the context of the exercise, we use such properties of the Gamma function to establish the recurrence relation \( \Gamma(x+1) = x \Gamma(x) \), which is a cornerstone in its practical applications, such as simplifying complex integrals or in the field of combinatorics where factorials are widely employed.