Question

Define the integrals \(I_{n}=\int_{-\infty}^{\infty} x^{2 n} e^{-x^{2}} d x .\) Noting that \(I_{0}=\sqrt{\pi}\) a. Find a recursive relation between \(I_{n}\) and \(I_{n-1} .\) b. Use this relation to determine \(I_{1}, I_{2}\), and \(I_{3}\). c. Find an expression in terms of \(n\) for \(I_{n}\).

Short answer

The recursive relation is \(I_{n} = 2nI_{n-1}\) using this result to the given \(I_{0} = \sqrt{\pi}\) we find \(I_{1} = \sqrt{\pi}\), \(I_{2} = 2\sqrt{\pi}\) and \(I_{3} = 12\sqrt{\pi}\). Hence, a generalized formula can be derived: \(I_{n} = n!(2^{n})\sqrt{\pi}\)

Step-by-step solution

Define The Recursive Relation

To find the recursive relation, integrating by parts is required. Using the formula \(\int u dv = uv - \int v du\), let \(u = x^{2n}\) and \(dv = e^{-x^{2}} dx\). Calculate \(du\) and \(v\), and substitute these values into the integration by parts formula. This will help to establish the relationship between \(I_{n}\) and \(I_{n-1}\).

Use The Recursive Relation

Once the recursive relation from step one has been found, it should be used to determine the values of \(I_{1}, I_{2}\), and \(I_{3}\). To do this, start with the provided value of \(I_{0}\) which is equal to \(\sqrt{\pi}\) and then apply the recursive relation iteratively for \(n = 1, 2, 3\).

Find a General Expression

In this step, upon evaluation of the values of \(I_{1}, I_{2}\) and \(I_{3}\), observe the pattern and formulate a general term \(I_{n}\) based on the pattern observed. This will involve using the concept of mathematical induction and pattern recognition.

Key concepts

Recursive Relation

In integrals related to mathematical physics, "recursive relation" is an important method used to solve problems systematically. A recursive relation allows solving for an integral or a function based on its previous values. This is very powerful because it reduces the problem's complexity step by step.

In the exercise, we use the recursive relation by expressing the integral \( I_n \) in terms of the prior integral \( I_{n-1} \). By applying integration by parts, which we'll discuss shortly, the integral involving \( x^{2n} e^{-x^2} \) is related to the one involving \( x^{2(n-1)} e^{-x^2} \). This simplifies the computation significantly as each integral can be expressed in terms of the simpler previous integral.

Remember, recursive relations turn a daunting problem into a series of smaller, more manageable problems.

Integration by Parts

The technique of 'integration by parts' is a vital tool in calculus for handling specific types of integrals where the standard integration rules do not directly apply.

When dealing with a product of functions, such as \( x^{2n} e^{-x^2} \), integration by parts is your friend. The formula to remember is \( \int u \, dv = uv - \int v \, du \). For the problem at hand, choosing \( u = x^{2n} \) and \( dv = e^{-x^2} dx \) helps us find a way to link our target integral, \( I_n \), with \( I_{n-1} \), through simplifying \( du \) and \( v \) correspondingly.

This strategic breakdown transforms the original complex integral into something more manageable while leveraging previously solved integral forms.

Mathematical Induction

Mathematical induction is an elegant method for verifying the general expression or property, especially when it involves recursive processes and series of numbers. This principle proves that if something is true for a base case and the transition from one case to the next is valid, then it is true for all subsequent cases.

In our exercise, after determining values for \( I_1, I_2, \) and \( I_3 \) using the recursive relation, induction enables us to establish a general formula for \( I_n \). The process involves showing \( I_n \) based on the hypothesis that \( I_k \) is true for all smaller values \( k \), thereby validating the pattern we've seen is universally applicable.

This not only reinforces our results but helps reveal the implicit structure linking all the integrals together.