The integral int_{0}^{1} frac{1}{7^{left[frac | vedclass

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Question

(A) Let $I = \int_{0}^{1} 7^{-\left[\frac{1}{x}\right]} dx$. Let $n = \left[\frac{1}{x}\right]$,then $n \le \frac{1}{x} < n+1$,which implies $\frac{1}

Vedclass · Accepted Answer

$1+6 \log _{e}\left(\frac{6}{7}\right)$

Vedclass · Answer

(A) Let $I = \int_{0}^{1} 7^{-\left[\frac{1}{x}\right]} dx$. Let $n = \left[\frac{1}{x}\right]$,then $n \le \frac{1}{x} As $x$ goes from $0$ to $1$,$n$ goes from $\infty$ to $1$.$I = \sum_{n=1}^{\infty} \int_{\frac{1}{n+1}}^{\frac{1}{n}} 7^{-n} dx = \sum_{n=1}^{\infty} 7^{-n} \left( \frac{1}{n} - \frac{1}{n+1} \right) = \sum_{n=1}^{\infty} \frac{1}{7^n n} - \sum_{n=1}^{\infty} \frac{1}{7^n (n+1)}$.Using the expansion $-\ln(1-x) = \sum_{n=1}^{\infty} \frac{x^n}{n}$,we have $\sum_{n=1}^{\infty} \frac{(1/7)^n}{n} = -\ln(1 - 1/7) = -\ln(6/7) = \ln(7/6)$.For the second part,$\sum_{n=1}^{\infty} \frac{1}{7^n (n+1)} = 7 \sum_{n=1}^{\infty} \frac{(1/7)^{n+1}}{n+1} = 7 \sum_{k=2}^{\infty} \frac{(1/7)^k}{k} = 7 [-\ln(1 - 1/7) - 1/7] = 7 [\ln(7/6) - 1/7] = 7 \ln(7/6) - 1$.Thus,$I = \ln(7/6) - (7 \ln(7/6) - 1) = 1 - 6 \ln(7/6) = 1 + 6 \ln(6/7)$.

The integral $\int_{0}^{1} \frac{1}{7^{\left[\frac{1}{x}\right]}} dx$,where $[.]$ denotes the greatest integer function,is equal to:

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