Question

For the following exercises, consider the gamma function given by \(\Gamma(a)=\int_{0}^{\infty} e^{-y} y^{a-1} d y\)Show that \(\Gamma(a)=(a-1) \Gamma(a-1)\).

Short answer

The gamma function satisfies \( \Gamma(a) = (a-1)\Gamma(a-1) \) by integration by parts.

Step-by-step solution

Understand the Problem Statement

The problem requires us to prove the functional equation \( \Gamma(a) = (a-1)\Gamma(a-1) \). This is an important property of the gamma function.

Apply Integration by Parts

Consider the integral \( \Gamma(a) = \int_{0}^{\infty} e^{-y} y^{a-1} dy \). We can apply integration by parts, choosing \( u = y^{a-1} \) and \( dv = e^{-y} dy \). Then we have \( du = (a-1)y^{a-2} dy \) and \( v = -e^{-y} \).

Evaluate the Integration by Parts Formula

The integration by parts formula is \( \int u\, dv = uv - \int v\, du \). Plugging in the values, we get:\[\int_{0}^{\infty} e^{-y} y^{a-1} dy = \left[ -e^{-y} y^{a-1} \right]_{0}^{\infty} + (a-1)\int_{0}^{\infty} e^{-y} y^{a-2} dy.\]The term \( \left[ -e^{-y} y^{a-1} \right]_{0}^{\infty} \) evaluates to zero because the exponential vanishes at both 0 and infinity.

Simplify the Resulting Expression

Now we are left with \(\Gamma(a) = (a-1) \int_{0}^{\infty} e^{-y} y^{a-2} dy \). The remaining integral is \(\int_{0}^{\infty} e^{-y} y^{a-2} dy = \Gamma(a-1)\).

Establish the Required Relationship

Substitute the integral for \(\Gamma(a-1)\) back into the equation: \[ \Gamma(a) = (a-1)\Gamma(a-1) \]. This completes the proof of the required functional equation.

Key concepts

Integration by Parts

Integration by parts is a technique used to solve definite and indefinite integrals, especially when they involve products of functions. This method is derived from the product rule of differentiation and helps simplify complex integrals into more manageable parts. To apply integration by parts, we use the formula: \[ \int u \, dv = uv - \int v \, du \]Here is a simple step-by-step on how to use this method effectively:

• Choose which parts of the integral to set as \( u \) (a function) and \( dv \) (a differential).
• Differentiating \( u \) gives \( du \), and integrating \( dv \) gives \( v \).
• Substitute these into the integration by parts formula.
• Solve the resulting integrals and simplify the expression.

In our problem, we set \( u = y^{a-1} \) and \( dv = e^{-y} dy \). Differentiating and integrating these gives:- \( du = (a-1) y^{a-2} dy \)- \( v = -e^{-y} \)By substituting these into the formula, we can transform and solve the original integral, illustrating the power and utility of integration by parts.

Functional Equation

A functional equation is an equation in which the unknowns are functions rather than just simple values. These equations show how functions relate to each other, offering deep insights into their properties.In mathematics, the gamma function's functional equation \( \Gamma(a) = (a-1)\Gamma(a-1) \) is particularly significant. This equation relates the gamma function at one point to its value at another, providing a recursive technique to compute gamma functions for various arguments. The beauty of functional equations lies in their ability to define a complex system in terms of simpler, often recursive relationships. Understanding these can help solve various mathematical and applied problems efficiently.In this case, to show the functional equation's validity, we use integration by parts to transform expressions involving gamma functions into a simplified, equivalent form. This process demonstrates how functional equations can reveal hidden connections between different mathematical objects.

Definite Integral

A definite integral involves computing the area under the curve for a given function between specific limits or bounds. It has practical applications in physics, engineering, and probability, among other fields.Mathematically, it is represented as:\[ \int_{a}^{b} f(x) \, dx \]This notation represents the area enclosed by the graph of \( f(x) \) from \( x = a \) to \( x = b \). It can also model accumulated quantities, like total distance traveled or the cumulative distribution function in probability.In our exercise, we focus on the definite integral of the form \[ \Gamma(a) = \int_{0}^{\infty} e^{-y} y^{a-1} dy \]This integral is central to defining the gamma function itself. The infinite upper limit suggests an extension beyond typical bounds, showcasing the versatility of definite integrals in defining generalized functions like \( \Gamma(a) \). By managing limits carefully and applying techniques such as integration by parts, we can evaluate seemingly difficult definite integrals with precision. This is key in proving relationships like the one shown in the gamma function's identity.