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

The Correct Option is A

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

\(\frac{−1}{x^2}dx=dt\)

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

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

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

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

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

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

=\(1+6\log\frac{6}{7}\)