Question

The integral $\underset{0}{\overset{1}{\int }}\frac{1}{7^{[\frac{1}{x}}]}dx$, where $[.]$ denotes the greatest integer function is equal to

• A. $1+6{\log }_e\left(\frac{6}{7}\right)$
• B. $1-6{\log }_e\left(\frac{6}{7}\right)$
• C. ${\log }_e\left(\frac{7}{6}\right)$
• D. $1-7{\log }_e\left(\frac{6}{7}\right)$

Answer 1

Answer: (A)

$\underset{0}{\overset{1}{\int }}\frac{1}{7^{\left[\frac{1}{x}\right]}}dx=-\underset{1}{\overset{0}{\int }}\frac{1}{7^{\left[\frac{1}{x}\right]}}dx$

$=(-1)\left[\underset{1}{\overset{1/2}{\int }}\frac{1}{7}dx+\underset{1/2}{\overset{1/3}{\int }}\frac{1}{7^2}dx+\underset{1/3}{\overset{1/4}{\int }}\frac{1}{7^3}dx+\dots \dots \infty \right]=\left(\frac{1}{7}+\frac{1}{2\cdot 7^2}+\frac{1}{3\cdot 7^3}+\dots \infty \right)-\left(\frac{1}{7\cdot 2}+\frac{1}{7^2\cdot 3}+\frac{1}{7^2\cdot 4}\dots \infty \right)$
$=-\ln \left(1-\frac{1}{7}\right)-7\left(\frac{1}{7^2\cdot 2}+\frac{1}{7^3\cdot 3}+\frac{1}{7^4\cdot 4}+\dots \dots \infty \right)$
$=\left[\text{as}\ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}\dots \infty \right]$
$[$ as $\ln (1-x)=-\left(x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}\dots \infty \right)]$
$=6\ln \frac{6}{7}-7\left(-\ln \left(1-\frac{1}{7}\right)-\frac{1}{7}\right)$
$=6\ln \frac{6}{7}+1$