Problem 99 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(\int_{0}^{\infty} e^{-a x} d x=\frac{1}{a}\), then \(\int_{0}^{\infty} x^{n} e^{-a x} d x\) is (A) \(\frac{(-1)^{n} n !}{a^{n+1}}\) (B) \(\frac{(-1)^{n}(n-1) !}{a^{n}}\) (C) \(\frac{n !}{a^{n+1}}\) (D) None of these

Step-by-Step Solution

Answer

The answer is (C) \(\frac{n !}{a^{n+1}}\).

Step 1: Understand the Problem

We are given that \(\int_{0}^{\infty} e^{-a x} \, dx=\frac{1}{a}\). We need to evaluate \(\int_{0}^{\infty} x^{n} e^{-a x} \, dx\). This is a type of improper integral often encountered in calculus, specifically a Laplace transform.

Step 2: Identify Use of Formula

We recognize that \(\int_{0}^{\infty} x^{n} e^{-a x} \, dx\) can be solved using the Gamma function property: \(\Gamma(n+1) = \int_{0}^{\infty} x^{n} e^{-x} \, dx = n!\). For \(e^{-a x}\), we substitute \(ax = y\), implying \(x = \frac{y}{a}, \, dx = \frac{dy}{a}\).

Step 3: Substitute Variables

Change variables using \(y = ax\). Then \(x = \frac{y}{a}\) and \(dx = \frac{dy}{a}\). This implies: \[\int_{0}^{\infty} x^{n} e^{-a x} \, dx = \int_{0}^{\infty} \left(\frac{y}{a}\right)^{n} e^{-y} \left(\frac{dy}{a}\right)\].

Step 4: Simplify Expression

Factor out everything that doesn't depend on \(y\): \[= \frac{1}{a^{n+1}} \int_{0}^{\infty} y^{n} e^{-y} \, dy\]. Now we see \(\int_{0}^{\infty} y^{n} e^{-y} \, dy = n!\).

Step 5: Apply Gamma Function Result

Since \(\int_{0}^{\infty} y^{n} e^{-y} \, dy = n!\), substitute it back into the integral: \[= \frac{1}{a^{n+1}} n!\].

Step 6: Choose the Correct Option

The expression \(\frac{n!}{a^{n+1}}\) matches option (C), confirming the solution to the given problem.

Key Concepts

Improper integral

An improper integral is a type of integral where the limits of integration are either infinite or the integrand becomes infinite at some point within the limits. This type of integral is essential in calculus to compute areas under curves that approach infinity. Frequently, improper integrals extend from a specified point to infinity, such as \( \int_{0}^{\infty} \).
To handle improper integrals, we often need to consider limits. Imagine you need to find the integral from 0 to infinity of \( e^{-ax} \). This requires examining the limit as the upper boundary tends to infinity:
- Set the integral with a variable such as \( t \) taking the place of infinity.
- Calculate the integral from 0 to \( t \) and let \( t \to \infty \).

In our exercise, we evaluate \( \int_{0}^{\infty} x^n e^{-ax} \, dx \). It's improper because the upper limit is infinite. The known result \( \int_{0}^{\infty} e^{-ax} \, dx = \frac{1}{a} \) serves as a building block for determining more complex integrals by eliminating infinite boundaries.

Laplace transform

The Laplace transform is a powerful integral transform used to convert functions of time into functions of a complex variable. This method is used extensively for solving differential equations and analyzing systems.
The Laplace transform of a function \( f(t) \) is given by:\[L\{f(t)\} = \int_{0}^{\infty} e^{-st} f(t) \, dt\]Where \( s \) is a complex number parameter.

In the context of the given exercise, we recognize a resemblance to the Laplace transform. The integral \( \int_{0}^{\infty} e^{-ax} \, dx \) closely relates to the Laplace transform with \( s = a \). This technique transforms the exponential decay function, helping simplify our problem into manageable forms. It's notable that the Laplace transform greatly facilitates handling improper integrals by providing a tool that reshapes continuous time-domain functions into a format suitable for algebraic manipulations.

Variable substitution

Variable substitution, also known as change of variables, simplifies complex integral calculations by transforming the integrand into a more straightforward form. This technique is crucial for solving integrals involving difficult functions or variables.
To perform a variable substitution:

• Select a new variable (e.g., \( y = ax \)) to substitute into the integral.
• Express the old variables in terms of the new variable (here, \( x = \frac{y}{a} \) and \( dx = \frac{dy}{a} \)).
• Rewrite the integral using the new variable setup.

The exercise applies this method when switching from \( x \) to \( y = ax \), allowing the removal of the parameter \( a \) from the exponent. As a result, the integral simplifies into a recognizable gamma function form:\[\int_{0}^{\infty} \left(\frac{y}{a}\right)^{n} e^{-y} \left(\frac{dy}{a}\right)\]This approach facilitates integral computation, providing a route to analytically solve otherwise challenging problems involving exponential functions.