Problem 33 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

Either evaluate the given improper integral or show that it diverges. $$ \int_{-\infty}^{\infty} x^{3} e^{-x^{2}} d x $$

Step-by-Step Solution

Answer

The improper integral converges and its value is 1.

Step 1: Break the integral into two parts

Evaluate the improper integral by splitting it into two integrals over the intervals \( -\infty \) to 0 and 0 to \( \infty \).\[ I = \int_{-\infty}^{\infty} x^{3} e^{-x^{2}} \, dx = \int_{-\infty}^{0} x^{3} e^{-x^{2}} \, dx + \int_{0}^{\infty} x^{3} e^{-x^{2}} \, dx \]

Step 2: Evaluate the first integral

For \(x \in (-\infty, 0]\): Let \( x = -t \) where \( t > 0 \). Then, \( dx = -dt \).\[ \int_{-\infty}^{0} x^{3} e^{-x^{2}} \, dx = \int_{\infty}^{0} (-t)^3 e^{-t^{2}} (- dt) = \int_{\infty}^{0} -t^3 e^{-t^{2}} \, (-dt) = \int_{0}^{\infty} t^3 e^{-t^{2}} \, dt \]

Step 3: Evaluate the second integral

For \( x \in [0, \infty) \): \[ \int_{0}^{\infty} x^{3} e^{-x^{2}} \, dx \]

Step 4: Combine and Simplify

Both integrals are the same. Combine and simplify:\[ I = 2 \int_{0}^{\infty} x^{3} e^{-x^{2}} \, dx \]

Step 5: Substitution

Use the substitution \( u = x^2 \), thus \( du = 2x \, dx \) or \( x \, dx = \frac{1}{2} du \). After substituting:\[ I = 2 \int_{0}^{\infty} x^{3} e^{-x^{2}} \, dx = 2 \int_{0}^{\infty} x^2 \, x \, e^{-x^2} \, dx = 2 \int_{0}^{\infty} x^2 \, e^{-x^2} ( x \, dx) = \int_{0}^{\infty} x^2 \, e^{ -u } \frac{1}{2} \, du = \int_{0}^{\infty} \frac{1}{2} \, u \, e^{ -u } \, du \]

Step 6: Solve the gamma function

Multiply by 2: \[ I = 2 \times \frac{1}{2} = 1 \]

Key Concepts

Improper Integrals

Improper integrals are a special type of integrals where either the limits of integration are infinite, or the function being integrated has an infinite discontinuity within the bounds of integration. In this exercise, we evaluate the integral of the function \( x^3 e^{-x^2} \) over all real numbers, which inherently requires evaluating from \( -\function -- to \ to \ : I = localsystem I = localsystem I dx$. Clearly, the break within the Android is the - and . This transforms it into _although_ another _(' in they cases_). Other times, indefinite evaluation is within the function. Not clear type.\)