Problem 99 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(\int_{0}^{\infty} e^{-\alpha x} d x=\frac{1}{a}\), 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

Option (C) \(\frac{n!}{a^{n+1}}\).

Step 1: Recognize the Integral Formula

We are given that \( \int_{0}^{\infty} e^{-\alpha x} \, dx = \frac{1}{a} \). This is a standard exponential integral formula where \( \int_{0}^{\infty} e^{-ax} \, dx = \frac{1}{a} \). Therefore, \( \alpha = a \).

Step 2: Understand the New Integral Function

Now, we need to solve \( \int_{0}^{\infty} x^{n} e^{-a x} \, dx \). This is a form of the gamma function integral, \( \int_{0}^{\infty} x^{n} e^{-ax} \, dx = \frac{n!}{a^{n+1}} \).

Step 3: Match the Result

Comparing the integral result \( \frac{n!}{a^{n+1}} \) with the options provided, we see that it matches option (C): \( \frac{n!}{a^{n+1}} \).

Key Concepts

Exponential Integral

The Exponential Integral is an important concept in calculus, particularly when dealing with integrals where the exponent is linear in form. Understand that when you see the expression \( \int_{0}^{\infty} e^{-\alpha x} \, dx \), it's a typical exponential function integrated over an infinite range. In the given exercise, it's shown that this standard integral evaluates to \( \frac{1}{a} \) given the parameter \( \alpha = a \). This is customarily due to the behavior of the exponential function, where the exponential function quickly approaches zero as \( x \) increases to infinity, making the definite integral approach \( \frac{1}{a} \).
It is important to remember that exponential integrals often result from integrating functions of the form \( e^{-bx} \), given its unique property of rapid decay as \( x \) increases, ensuring convergence of the integral for positive values of \( b \).
Recognizing these patterns is key in solving problems where exponential integrals are involved.

Integration Techniques

Integration techniques often include substitution, integration by parts, or seeing patterns related to known integral forms. Here, for \( \int_{0}^{\infty} x^{n} e^{-a x} \, dx \), this integral can be solved by utilizing the form of the Gamma function integral. Remember, the technique of recognizing this type of integral form simplifies the solution process because it directly relates it to the known Gamma function result.
The Gamma function \( \Gamma(n+1) \) precisely equals \( n! \) (factorials of \( n \)), hence for these types of exponential integration problems, it pays off to notice that \( \int_{0}^{\infty} x^{n} e^{-a x} \, dx = \frac{n!}{a^{n+1}} \). This method is a classic example of when recognizing a formula quickly reduces the amount of algebraic computations required, making it a highly efficient technique.
Such understanding can save time in complex integration scenarios and is worth remembering during integral calculus studies.

Definite Integrals

Definite integrals are a way of calculating the area under a curve from one point to another, often from maintenance of a range. In this scenario, tackling a definite integral from \( 0 \) to \( \infty \) allows us to precisely evaluate the influence of a function over an infinite interval. In mathematical physics and engineering, such integrals are quite common.
In our exercise, both given integrals are examples of definite integrals, evaluated from zero to infinity. This is a crucial aspect as it deals specifically with infinite expansions, allowing certain integrals to resolve into known values. Here, the integral \( \int_{0}^{\infty} e^{-ax} \, dx \) results in \( \frac{1}{a} \) because it defines a finite area below the curve of an exponential function.
Understanding how to evaluate definite integrals, especially on infinite intervals, greatly enhances one's ability to solve problems where these mathematical concepts apply, such as in probability theory where similar integrals define distribution functions.