Question

Find the value of _limit ( tⁿ / e^t ) (t tends to infinity) and then prove that

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∫ (tⁿ / e^t ) dt = n factorial

limit of integration is from 0 to infinity

Answer 1

$$\begin{gathered} case\left(1\right):whennisnegativeinteger.Then,letn=-m,m\in N \\ So, \\ \underset{t\to \infty }{\lim }\frac{t^n}{e^t}=\underset{t\to \infty }{\lim }\frac{t^{-m}}{e^t} \\ =\underset{t\to \infty }{\lim }\frac{1}{e^t{t}^m} \\ =\frac{1}{\infty } \\ =0 \end{gathered}$$
$$\begin{gathered} case\left(2\right):whenn=0 \\ So, \\ \underset{t\to \infty }{\lim }\frac{t^n}{e^t}=\underset{t\to \infty }{\lim }\frac{t^0}{e^t} \\ =\underset{t\to \infty }{\lim }\frac{1}{e^t} \\ =\frac{1}{\infty } \\ =0 \end{gathered}$$
$$\begin{gathered} case\left(3\right):whennispositiveinteger. \\ So, \\ \underset{t\to \infty }{\lim }\frac{t^n}{e^t}=\underset{t\to \infty }{\lim }\frac{t^n}{e^t} \\ =\underset{t\to \infty }{\lim }\frac{n!}{e^t}\left[ByL\text{'}Hospital\text{'}srule\right] \\ =0 \\ Hence,\underset{t\to \infty }{\lim }\frac{t^n}{e^t}=0forallvaluesofn. \end{gathered}$$
Now,
${\int }_0^{\infty }\frac{t^n}{e^t}dt={\int }_0^{\infty }tne-tdt$
$={\int }_0^{\infty }t\left(n+1\right)-1e-tdt$
$$\begin{gathered} =gammafunction\left(n+1\right) \\ =n! \end{gathered}$$