Question

In kinetic theory we have to evaluate integrals of the form \(I=\int_{0}^{\infty} t^{n} e^{-a t^{2}} d t .\) Given that \(\int_{0}^{\infty} e^{-a t^{2}} d t=\frac{1}{2} \sqrt{\pi / a},\) evaluate \(I\) for \(n=2,4,6, \cdots, 2 m.\)

Short answer

For \ n = 2m, \ \int_{0}^{\infty} t^{2m} e^{-a t^{2}} d t=\frac{(2m-1)!}{2^{m+1}a^{m+0.5} m!} \sqrt{\pi}.\)

Step-by-step solution

Recognize the Given Integral

The integral given is \(I=\int_{0}^{\infty} t^{n} e^{-a t^{2}} d t.\) We are provided with the specific integral \(\int_{0}^{\infty} e^{-a t^{2}} d t=\frac{1}{2} \sqrt{\pi / a}.\)

Use a Substitution Method

For convenience, let’s introduce a substitution method. Let \(u = \sqrt{a} t.\) This implies \(d u = \sqrt{a} d t.\) Substituting these into the integral, we get \(I =\int_{0}^{\infty} (\frac{u}{\sqrt{a}})^{n} e^{-u^{2}} * \frac{d u}{\sqrt{a}}.\)

Simplify the Integral

Simplify the integral based on the substitution. The integral transforms to \(I =\frac{1}{a^{\frac{n+1}{2}}}\int_{0}^{\infty} u^{n} e^{-u^{2}} d u.\)

Recognize the Gamma Function Integral

Identify the Gamma function integral formula: \(\int_{0}^{\infty} x^{n} e^{-x^{2}} d x =\frac{1}{2} \Gamma\left(\frac{n+1}{2}\right).\) Rewriting in terms of our variables, \(I = \frac{1}{2 a^{\frac{n+1}{2}}}\Gamma\left(\frac{n+1}{2}\right). \)

Evaluate for Even n

For even n (i.e., n = 2m), the result is similar. Substitute n = 2m in the previous step to obtain \(I = \frac{1}{2 a^{m+1}} \Gamma(m + \frac{1}{2}).\) Moving forward, recognize \(\Gamma(m + 0.5) = \frac{(2m-1)!}{2^{m} {m}!} \sqrt{\pi}.\)

Final Integral Result

By substitution, the integration result for n = 2m is \(\int_{0}^{\infty} t^{2m} e^{-a t^{2}} d t =\frac{(2m-1)!}{2^{m+1}a^{m+0.5} m!} \sqrt{\pi}.\)

Key concepts

Substitution Method in Integrals

The substitution method is a helpful technique for solving integrals. It involves changing the variable of integration to simplify the integral.
This can be particularly useful when you're dealing with complex functions. Here's how it works in the context of kinetic theory integrals:
Let's start with the integral \(I = \int_{0}^{\infty} t^{n} e^{-a t^{2}} d t\).
We can use a simple substitution to make it easier to solve.

• Let \(u = \sqrt{a} t\).
• This implies \(d u = \sqrt{a} d t\).
  By substituting these into the integral, we get \(I =\int_{0}^{\infty} (\frac{u}{\sqrt{a}})^{n} e^{-u^{2}} * \frac{d u}{\sqrt{a}}\).
• Now, simplify the integral based on the substitution.
  This transformation is powerful because it reduces a seemingly complicated integral into a more manageable form.
  Using substitution in integrals helps reduce the complexity of solving them, making it easier to find a solution. It is a vital technique when dealing with integrals, especially those arising in advanced topics like kinetic theory.
  

Gamma Function

The Gamma function, denoted as \(\Gamma(x)\), plays a crucial role in various areas of mathematics, including integrals and kinetic theory.
It extends the concept of factorial (\(n!\)) to non-integer values.\

• For positive integers, \(\Gamma(n) = (n-1)!\).
• The Gamma function is defined as \(\Gamma(z) = \int_{0}^{\infty} t^{z-1} e^{-t} d t.\)
  One key identity involving the Gamma function and relevant to our context is \(\int_{0}^{\infty} x^{n} e^{-x^{2}} d x = \frac{1}{2} \Gamma\left(\frac{n+1}{2}\right).\)
  In our solution, this identity allowed the integral to be written in terms of the Gamma function. This simplification is significant because it lets us express complex integrals in a more tractable form.
  Knowing how to use the Gamma function and its properties is essential for solving higher-order integrals and other advanced topics in mathematics.
  

Higher-Order Integrals

Higher-order integrals refer to integrals involving functions raised to higher powers or involving products of several functions.
These integrals are more complex than basic integrals, often requiring special techniques or additional knowledge to solve.
In kinetic theory, you often encounter integrals of the form \(I =\int_{0}^{\infty} t^{n} e^{-a t^{2}} d t\), especially for even values of \(n\) (i.e., \(n=2m\)).

• Identifying patterns or specific transformations, like substitutions, can simplify the process.
• In our example, we substitute and simplify the integral to reach a form involving the Gamma function.
  For even \(n\), the integral \(I = \frac{1}{2 a^{m+1}} \Gamma(m + \frac{1}{2})\) gives an elegant result.
  Expressing the solution using known formulas and identities, such as those involving the Gamma function, is a crucial step in solving higher-order integrals.
  This methodical approach aids students in tackling more challenging integrals, emphasizing the importance of understanding and utilizing advanced mathematical concepts.