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  4. If ∫ limits∞0e-axdx=(1/a), then ∫ limits∞0xn e-axdx is

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Q. If $\int\limits^{\infty}_{{0}}e^{-ax}dx=\frac{1}{a},$ then $\int\limits^{\infty}_{{0}}x^n\,e^{-ax}dx$ is

Integrals

Solution:

Let $I=\int\limits^{\infty}_{{0}}x^n\,e^{-ax}=$ $\left[x^{n}\cdot \frac{e^{-ax}}{-a}\right]^{^{\infty}}_{_{_0}}-$ $\int\limits^{\infty}_{{0}}nx^{n-1}\cdot \frac{e^{-ax}}{-a}dx$
$=-\frac{1}{a}$ $\displaystyle \lim_{x \to \infty} $$=\frac{x^{n}}{e^{ax}}+\frac{n}{a}I_{n-1}$
$\therefore I_{n}=\frac{n}{a}I_{n-1}\,\left[\because \displaystyle \lim_{x \to \infty} \frac{x^{n}}{e^{ax}}=0\right]$
$=\frac{n}{a}\cdot \frac{n-1}{a}I_{n-2}=\frac{n\left(n-1\right)\left(n-2\right)}{a^{3}}I_{n-3}$
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$=\frac{n!}{a^{n}}$$\int\limits^{\infty}_{{0}}$$e^{-ax}dx=\frac{n!}{a^{n+1}}$