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Show that $\int_{0}^{\infty}\frac{x^7}{7^x}dx=\frac{7!}{(log\,7)^8}$ .

written 8.2 years ago by teamques10 ★ 70k

•   modified 6.0 years ago

engineering mathematics

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written 8.0 years ago by teamques10 ★ 70k

•   modified 8.0 years ago

Put,

$7^x=e^t \hspace{0.3cm}=\gt\hspace{0.3cm}x=\frac{x}{log7}\hspace{0.3cm}=\gt\hspace{0.3cm}dx=\frac{dt}{log7}$

When,

$\,\,x=0,t=0,x=\infty,t=\infty,$

$\int_{0}^{\infty} (\frac{t}{log7})^7 e^{-t} \frac{dt}{log7}=\frac{1}{(log7)^{8}}\int_{0}^{\infty}e^{-t}t^7dt \\ \hspace{3cm}=\frac{1}{(log7)^8} \sqrt{8}=\frac{7!}{(log7)^8}$

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