Question

If ${\int }_0^{\infty }{e}^{-ax}dx=\frac{1}{a}$ then ${\int }_0^{\infty }{x}^n{e}^{-ax}dx$ is

• A. $({(-1)}^{n}n!)/{\left(a\right)}^{n+1}$
• B. $({(-1)}^n(n-1)!)/{\left(a\right)}^n$
• C. $n/(a^n+1)$
• D. None of these

Answer 1

The correct option is D None of these

Given

${\int }_0^{\infty }{e}^{-ax}dx=\frac{1}{a}$. Differentiating successively with respect to the variable $a$ we get

${\int }_0^{\infty }{e}^{-ax}(-x)dx=\frac{-1}{a^2}$

${\int }_0^{\infty }{e}^{-ax}(-x{)}^{2}dx=\frac{2}{a^3}$

${\int }_0^{\infty }{e}^{-ax}(-x{)}^{3}dx=\frac{-3!}{a^4}$

Hence differentiating n times we get

${\int }_0^{\infty }{e}^{-ax}(-x{)}^{n}dx=\frac{(-1{)}^{n}n!}{a^{n+1}}$

dividing by $(-1{)}^n$on both sides we get

${\int }_0^{\infty }{e}^{-ax}{x}^{n}dx=\frac{n!}{a^{n+1}}$.

Hence the correct answer is option D none of these.