863 e- dx r=0
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Riemann Sum
Riemann Sums is a special type of approximation of the area under a curve by dividing it into multiple simple shapes like rectangles or trapezoids and is used in integrals when finite sums are involved. Figuring out the area of a curve is complex hence this method makes it simple. Usually, we take the help of different integration methods for this purpose. This is one of the major parts of integral calculus.
Riemann Integral
Bernhard Riemann's integral was the first systematic description of the integral of a function on an interval in the branch of mathematics known as real analysis.
Question
Find the following definite, improper,
![**Problem (3):**
Evaluate the integral
\[
\int_{x=0}^{+\infty} x^{863} e^{-x} \, dx = \underline{\hspace{2cm}}.
\]
---
**Explanation:**
In this problem, you are asked to evaluate an integral of the form:
\[
\int_{0}^{\infty} x^a e^{-x} \, dx,
\]
where \( a = 863 \) in this specific problem.
This integral is a Gamma function in disguise. The Gamma function \(\Gamma(n)\) is defined for \( n > 0 \) as:
\[
\Gamma(n) = \int_{0}^{\infty} x^{n-1} e^{-x} \, dx.
\]
By comparing the given integral to the Gamma function definition, we see that \( a = 863 \), so we can relate it to the Gamma function by shifting the exponent:
\[
\int_{x=0}^{+\infty} x^{863} e^{-x} \, dx = \Gamma(863+1).
\]
Thus, we can express the result in terms of the Gamma function:
\[
\Gamma(864).
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F239df334-5b03-4bc0-9001-d398335a6cd3%2Ff37c25ee-b51c-42fb-9580-95c945ab4162%2F35en78.png&w=3840&q=75)
Transcribed Image Text:**Problem (3):**
Evaluate the integral
\[
\int_{x=0}^{+\infty} x^{863} e^{-x} \, dx = \underline{\hspace{2cm}}.
\]
---
**Explanation:**
In this problem, you are asked to evaluate an integral of the form:
\[
\int_{0}^{\infty} x^a e^{-x} \, dx,
\]
where \( a = 863 \) in this specific problem.
This integral is a Gamma function in disguise. The Gamma function \(\Gamma(n)\) is defined for \( n > 0 \) as:
\[
\Gamma(n) = \int_{0}^{\infty} x^{n-1} e^{-x} \, dx.
\]
By comparing the given integral to the Gamma function definition, we see that \( a = 863 \), so we can relate it to the Gamma function by shifting the exponent:
\[
\int_{x=0}^{+\infty} x^{863} e^{-x} \, dx = \Gamma(863+1).
\]
Thus, we can express the result in terms of the Gamma function:
\[
\Gamma(864).
\]
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