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Mathematics Question on integral

If ∫0∞e−axdx=1a,\int\limits^{\infty}_{{0}}e^{-ax}dx=\frac{1}{a}, then ∫0∞xn e−axdx\int\limits^{\infty}_{{0}}x^n\,e^{-ax}dx is

A

(−1)nn!an+1\frac{\left(-1\right)^{n}n!}{a^{n+1}}

B

(−1)n(n−1)!an\frac{\left(-1\right)^{n}\left(n-1\right)!}{a^{n}}

C

n!an+1\frac{n!}{a^{n+1}}

D

None of these

Answer

n!an+1\frac{n!}{a^{n+1}}

Explanation

Solution

Let I=∫0∞xn e−ax=I=\int\limits^{\infty}_{{0}}x^n\,e^{-ax}= [xn⋅e−ax−a]0∞−\left[x^{n}\cdot \frac{e^{-ax}}{-a}\right]^{^{\infty}}_{_{_0}}- ∫0∞nxn−1⋅e−ax−adx\int\limits^{\infty}_{{0}}nx^{n-1}\cdot \frac{e^{-ax}}{-a}dx =−1a=-\frac{1}{a} lim⁡x→∞\displaystyle \lim_{x \to \infty} =xneax+naIn−1=\frac{x^{n}}{e^{ax}}+\frac{n}{a}I_{n-1} ∴In=naIn−1 [∵lim⁡x→∞xneax=0]\therefore I_{n}=\frac{n}{a}I_{n-1}\,\left[\because \displaystyle \lim_{x \to \infty} \frac{x^{n}}{e^{ax}}=0\right] =na⋅n−1aIn−2=n(n−1)(n−2)a3In−3=\frac{n}{a}\cdot \frac{n-1}{a}I_{n-2}=\frac{n\left(n-1\right)\left(n-2\right)}{a^{3}}I_{n-3} .................................................................................. .............................................................................. =n!an=\frac{n!}{a^{n}} ∫0∞\int\limits^{\infty}_{{0}} e−axdx=n!an+1e^{-ax}dx=\frac{n!}{a^{n+1}}