Question

Evaluate the integrals (a) \(\int_{0}^{\infty} x^{2} e^{-2 x^{2}} d x\), (b) \(\int_{0}^{\infty} \sqrt[4]{x} e^{-\sqrt{x}} d x\)

Short answer

(a) \( \frac{\sqrt{\pi}}{16} \), (b) \( \frac{3\sqrt{\pi}}{2} \)

Step-by-step solution

Recognize the Integral Form - Part (a)

Observe that the first integral \[ I = \int_{0}^{\infty} x^2 e^{-2x^2} \, dx \]resembles a Gaussian-type integral. Use the substitution \( u = \sqrt{2}x \) which gives \( du = \sqrt{2} \, dx \). This transforms the bounds from 0 to \( \infty \) and simplifies the integral form.

Substitute and Simplify - Part (a)

With the substitution \( u = \sqrt{2}x \), we have \( dx = \frac{du}{\sqrt{2}} \) and the integral becomes:\[ I = \int_{0}^{\infty} \left(\frac{u}{\sqrt{2}}\right)^2 e^{-u^2} \cdot \frac{du}{\sqrt{2}} \]Simplify to:\[ I = \frac{1}{4} \int_{0}^{\infty} u^2 e^{-u^2} \, du \]

Evaluate the Gaussian Integral - Part (a)

The integral \( \int_{0}^{\infty} u^2 e^{-u^2} \, du \) can be solved using the Gamma function. Use the result for Gaussian integrals, where:\[ \int_{0}^{\infty} t^n e^{-t} \, dt = \Gamma(n+1) \]Applying the change of variables \( t = u^2 \), so \( n = \frac{1}{2} \), we find:\[ \int_{0}^{\infty} u^2 e^{-u^2} \, du = \frac{\sqrt{\pi}}{4} \]

Calculate Final Value - Part (a)

Combine this result with the expression from Step 2:\[ I = \frac{1}{4} \cdot \frac{\sqrt{\pi}}{4} = \frac{\sqrt{\pi}}{16} \]

Recognize the Integral Form - Part (b)

Now consider the integral:\[ J = \int_{0}^{\infty} x^{1/4} e^{-\sqrt{x}} \, dx \]Perform the substitution \( v = \sqrt{x} \), thus \( x = v^2 \) and \( dx = 2v \, dv \), adjusting the limits from 0 to \( \infty \).

Substitute and Simplify - Part (b)

Replace in terms of \( v \):\[ J = \int_{0}^{\infty} (v^2)^{1/4} e^{-v} \cdot 2v \, dv = 2 \int_{0}^{\infty} v^{\frac{3}{2}} e^{-v} \, dv \]

Evaluate the Gamma Function Integral - Part (b)

This is a Gamma function integral for which \( n = \frac{3}{2} \). The Gamma function relation gives us:\[ \int_{0}^{\infty} v^n e^{-v} \, dv = \Gamma(n+1) \]Thus:\[ \int_{0}^{\infty} v^{\frac{3}{2}} e^{-v} \, dv = \Gamma\left(\frac{5}{2}\right) \]

Compute Gamma Function - Part (b)

The value of \( \Gamma\left(\frac{5}{2}\right) \) is:\[ \Gamma\left(\frac{5}{2}\right) = \frac{3 \sqrt{\pi}}{4} \]Therefore:\[ J = 2 \times \frac{3 \sqrt{\pi}}{4} = \frac{3 \sqrt{\pi}}{2} \]

Key concepts

Gaussian Integrals

Gaussian integrals, a staple in integral calculus, are specifically useful when dealing with functions of the form \( e^{-ax^2} \). These integrals are pivotal in many areas including probability, physics, and statistics.

• Gaussian integrals often morph after using specific substitutions that simplify the calculations. For example, a common substitution is turning the variable to \( u = \sqrt{a}x \), shifting the integral to a more recognizable Gaussian form.
• The basic Gaussian integral, \( \int_{-\infty}^{\infty} e^{-ax^2} \ dx \), evaluates to \( \sqrt{\frac{\pi}{a}} \). This is a foundation upon which more complex calculations rest.

Understanding Gaussian integrals not only simplifies certain integration cases but also lays the groundwork for comprehending more complex concepts in higher mathematics.

Gamma Function

The Gamma function extends the factorial concept to real and complex numbers. It is a crucial tool in various branches of mathematics and physics due to its properties and applications.

• Defined by \( \Gamma(n) = \int_{0}^{\infty} t^{n-1} e^{-t} dt \) for \( n > 0 \), this function generalizes the factorial such that \( \Gamma(n+1) = n\Gamma(n) \) and \( \Gamma(n+1) = n! \) for natural numbers.
• The application of the Gamma function to integrals frequently comes into play when variables are changed to fit the integral form of the Gamma function, as seen in transformations during calculations.

The Gamma function not only serves in simplifying integrals but is also vital in solving differential equations and analyzing probability distributions among other applications.

Integral Transforms

Integral transforms, such as the Fourier or Laplace transforms, convert functions into a form that is often easier to manipulate. These transforms are instrumental in solving differential equations and in the fields of engineering and data analysis.

• They enable the transformation of a function into the integral of the function, often over an infinite interval, thereby solving or simplifying problems that might be intractable in their original form.
• In integral calculus, recognizing when an integral can be simplified or transformed into a known integral type, like the Gamma function form, often saves computation time and effort.

The transformative power of these mathematical tools facilitates analysis across diverse scientific and engineering problems, making complex issues more tractable.