Question

The integral $∫_0^ 1 \frac{ 1}{7^{[\frac{1}{z}}]}dx$ where [⋅] denotes the greatest integer function, is equal to

• A. $1+6log_e(\frac{6}{7})$
• B. $1-6log_e(\frac{6}{7})$
• C. $log_e(\frac{7}{6})$
• D. $1-7log_e(\frac{6}{7})$

Answer 1

Answer: (A)

$1+6log_e(\frac{6}{7})$

Answer 2

$∫_0^ 1 \frac{ 1}{7^{[\frac{1}{z}}]}dx$, let $\frac{1}{x}=t$

$\frac{−1}{x^2}dx=dt$

=$∫_∞ ^1 \frac{1}{−t^27[t]}dt=∫_1 ^∞ \frac{1}{t^27[t]}dt$

=$∫_1^ 2 \frac{1}{7t^2}dt+∫_2 ^3 \frac{1}{7^2t^2}dt+…$

=$\frac{1}{7}\bigg[−\frac{1}{t}\bigg]_1^ 2+\frac{1}{7 ^2}\bigg[\frac{−1}{t}\bigg]_2^ 3+\frac{1}{7 ^3}\bigg[\frac{−1}{t}\bigg]_2^ 3+…$

=$∑_{n=1} ^∞\frac{1}{7_n}\bigg(\frac{1}{n}−\frac{1}{n+1}\bigg)$

=$∑_{n=1} ^∞\frac{(\frac{1}{7})^n}{n}−7∑_{n=1} ^∞\frac{(\frac{1}{7})^{n+1}}{n+1}$

=$−log\bigg(1−\frac{1}{7}\bigg)+7log\bigg(1−\frac{1}{7}\bigg)+1$

=$1+6\log\frac{6}{7}$