Key Concepts and Formulas

• Greatest Integer Function: For an integer $n$, $[x] = n$ for $n \le x < n+1$. This property is crucial for breaking down integrals involving the greatest integer function.
• Definite Integration over Intervals: If a function $f(x)$ is defined piecewise, the integral $\int_a^b f(x) dx$ can be split into a sum of integrals over the subintervals where the function's definition is consistent.
• Geometric Series: The sum of an infinite geometric series with first term $a$ and common ratio $r$ (where $|r|<1$) is given by $\frac{a}{1-r}$.
• Logarithm Properties: Specifically, $\log_e(a/b) = \log_e a - \log_e b$.

Step-by-Step Solution

The integral we need to evaluate is $I = \int\limits_0^1 {{1 \over {{7^{\left[ {{1 \over x}} \right]}}}}dx}$

Step 1: Analyze the integrand and the greatest integer function. The integrand involves $[1/x]$. Let's consider the behavior of $[1/x]$ as $x$ ranges from $0$ to $1$. When $x$ is very close to $0$ (e.g., $x \to 0^+$), $1/x \to \infty$, so $[1/x]$ can take large integer values. When $x = 1$, $1/x = 1$, so $[1/x] = 1$. As $x$ approaches $0$ from the right, the value of $[1/x]$ changes only when $1/x$ crosses an integer value. This happens when $x$ takes values of the form $1/n$ for integers $n$.

Let $k = [1/x]$. If $1 \le 1/x < 2$, then $[1/x] = 1$. This corresponds to $1/2 < x \le 1$. If $2 \le 1/x < 3$, then $[1/x] = 2$. This corresponds to $1/3 < x \le 1/2$. If $n \le 1/x < n+1$, then $[1/x] = n$. This corresponds to $1/(n+1) < x \le 1/n$.

The integral is from $0$ to $1$. As $x \to 0^+$, $1/x \to \infty$. This means $[1/x]$ can take infinitely many integer values. We can split the integral into a sum of integrals over intervals where $[1/x]$ is constant. The intervals are of the form $(1/(n+1), 1/n]$. The integral from $0$ to $1$ can be written as a sum: $I = \int_0^1 \frac{1}{7^{[1/x]}} dx$ Let's consider the intervals based on the value of $[1/x]$. When $x=1$, $[1/x]=1$. When $1/2 < x < 1$, $1 < 1/x < 2$, so $[1/x] = 1$. When $1/3 < x \le 1/2$, $2 \le 1/x < 3$, so $[1/x] = 2$. When $1/4 < x \le 1/3$, $3 \le 1/x < 4$, so $[1/x] = 3$. In general, for $1/(n+1) < x \le 1/n$, we have $n \le 1/x < n+1$, so $[1/x] = n$.

The integral can be written as: $I = \sum_{n=1}^{\infty} \int_{1/(n+1)}^{1/n} \frac{1}{7^{[1/x]}} dx$ In the interval $(1/(n+1), 1/n]$, $[1/x] = n$. So the integrand becomes $1/7^n$. $I = \sum_{n=1}^{\infty} \int_{1/(n+1)}^{1/n} \frac{1}{7^n} dx$

Step 2: Evaluate the integral over each subinterval. For each $n$, the integral is: $\int_{1/(n+1)}^{1/n} \frac{1}{7^n} dx = \frac{1}{7^n} \int_{1/(n+1)}^{1/n} 1 dx$ $= \frac{1}{7^n} [x]_{1/(n+1)}^{1/n}$ $= \frac{1}{7^n} \left( \frac{1}{n} - \frac{1}{n+1} \right)$ $= \frac{1}{7^n} \left( \frac{(n+1) - n}{n(n+1)} \right)$ $= \frac{1}{7^n} \left( \frac{1}{n(n+1)} \right)$

Step 3: Sum the results to find the total integral. Now we need to sum this expression from $n=1$ to $\infty$: $I = \sum_{n=1}^{\infty} \frac{1}{7^n} \cdot \frac{1}{n(n+1)}$ $I = \sum_{n=1}^{\infty} \frac{1}{n(n+1)7^n}$

This sum is not directly a simple geometric series. Let's try a different approach using substitution.

Alternative Approach: Substitution

Let $y = 1/x$. Then $x = 1/y$, and $dx = -1/y^2 dy$. When $x = 0^+$, $y \to \infty$. When $x = 1$, $y = 1$. So the integral becomes: $I = \int_{\infty}^{1} \frac{1}{7^{[y]}} \left(-\frac{1}{y^2}\right) dy$ $I = \int_{1}^{\infty} \frac{1}{y^2 7^{[y]}} dy$

Now, we split this integral into intervals based on $[y]$. For $1 \le y < 2$, $[y] = 1$. For $2 \le y < 3$, $[y] = 2$. For $n \le y < n+1$, $[y] = n$.

$I = \int_{1}^{2} \frac{1}{y^2 7^1} dy + \int_{2}^{3} \frac{1}{y^2 7^2} dy + \int_{3}^{4} \frac{1}{y^2 7^3} dy + \cdots$ $I = \sum_{n=1}^{\infty} \int_{n}^{n+1} \frac{1}{y^2 7^n} dy$ $I = \sum_{n=1}^{\infty} \frac{1}{7^n} \int_{n}^{n+1} \frac{1}{y^2} dy$

Step 4: Evaluate the integral of $1/y^2$. The integral of $1/y^2$ is $-1/y$. $\int_{n}^{n+1} \frac{1}{y^2} dy = \left[-\frac{1}{y}\right]_{n}^{n+1} = -\frac{1}{n+1} - \left(-\frac{1}{n}\right) = \frac{1}{n} - \frac{1}{n+1} = \frac{1}{n(n+1)}$

Step 5: Sum the resulting series. Substituting this back into the sum: $I = \sum_{n=1}^{\infty} \frac{1}{7^n} \left(\frac{1}{n(n+1)}\right)$ $I = \sum_{n=1}^{\infty} \frac{1}{n(n+1)7^n}$

We can use partial fraction decomposition for $\frac{1}{n(n+1)}$: $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$ So, $I = \sum_{n=1}^{\infty} \left(\frac{1}{n} - \frac{1}{n+1}\right) \frac{1}{7^n}$ $I = \sum_{n=1}^{\infty} \left(\frac{1}{n \cdot 7^n} - \frac{1}{(n+1) \cdot 7^n}\right)$

Let's rewrite the terms: For $n=1$: $\frac{1}{1 \cdot 7^1} - \frac{1}{2 \cdot 7^1}$ For $n=2$: $\frac{1}{2 \cdot 7^2} - \frac{1}{3 \cdot 7^2}$ For $n=3$: $\frac{1}{3 \cdot 7^3} - \frac{1}{4 \cdot 7^3}$ ...

This doesn't seem to be a telescoping sum directly. Let's try to relate it to a known series expansion.

Consider the Maclaurin series for $-\log(1-x)$: $-\log(1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \cdots = \sum_{n=1}^{\infty} \frac{x^n}{n}$ This is for terms of the form $x^n/n$. Our sum has terms of the form $1/(n \cdot 7^n)$.

Let's consider another series. Consider the integral of a geometric series. $\int_0^x t^{n-1} dt = \frac{x^n}{n}$ Consider the series $\sum_{n=1}^{\infty} \frac{x^n}{n(n+1)}$. We know that $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$. So, $\sum_{n=1}^{\infty} \frac{x^n}{n(n+1)} = \sum_{n=1}^{\infty} \left(\frac{x^n}{n} - \frac{x^n}{n+1}\right)$.

Let's consider the integral of the series $\sum_{n=1}^{\infty} x^{n-1} = \frac{1}{1-x}$ for $|x|<1$. Integrating term by term: $\int_0^x \sum_{n=1}^{\infty} t^{n-1} dt = \sum_{n=1}^{\infty} \int_0^x t^{n-1} dt = \sum_{n=1}^{\infty} \left[\frac{t^n}{n}\right]_0^x = \sum_{n=1}^{\infty} \frac{x^n}{n} = -\log(1-x)$. Also, $\int_0^x \frac{1}{1-t} dt = [-\log(1-t)]_0^x = -\log(1-x)$.

Consider the integral of $-\log(1-x)$: $\int_0^x -\log(1-t) dt$. Let $u = -\log(1-t)$, $dv = dt$. Then $du = \frac{1}{1-t} dt$, $v = t$. Using integration by parts: $[-t \log(1-t)]_0^x - \int_0^x \frac{t}{1-t} dt$ $= -x \log(1-x) - \int_0^x \frac{-(1-t)+1}{1-t} dt$ $= -x \log(1-x) - \int_0^x \left(-1 + \frac{1}{1-t}\right) dt$ $= -x \log(1-x) - [-t - \log(1-t)]_0^x$ $= -x \log(1-x) - (-x - \log(1-x))$ $= -x \log(1-x) + x + \log(1-x)$ $= x + (1-x)\log(1-x)$.

Now, let's look at our sum $I = \sum_{n=1}^{\infty} \frac{1}{n(n+1)7^n}$. Let $x = 1/7$. We are evaluating $\sum_{n=1}^{\infty} \frac{1}{n(n+1)} \left(\frac{1}{7}\right)^n$.

Consider the series expansion: $\sum_{n=1}^{\infty} \frac{x^n}{n(n+1)} = \sum_{n=1}^{\infty} \left(\frac{x^n}{n} - \frac{x^n}{n+1}\right)$ $= \sum_{n=1}^{\infty} \frac{x^n}{n} - \sum_{n=1}^{\infty} \frac{x^n}{n+1}$ $= -\log(1-x) - \left(\frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \cdots \right)$ The second sum is $\left(\sum_{n=1}^{\infty} \frac{x^n}{n}\right) - x = -\log(1-x) - x$. So, $\sum_{n=1}^{\infty} \frac{x^n}{n(n+1)} = -\log(1-x) - (-\log(1-x) - x) = x$. This is incorrect.

Let's try integrating the geometric series in a different way. Consider the series $\sum_{n=1}^{\infty} x^n = \frac{x}{1-x}$ for $|x|<1$. Integrate from $0$ to $y$: $\int_0^y \sum_{n=1}^{\infty} t^n dt = \sum_{n=1}^{\infty} \int_0^y t^n dt = \sum_{n=1}^{\infty} \frac{y^{n+1}}{n+1}$. Also, $\int_0^y \frac{t}{1-t} dt = \int_0^y \frac{-(1-t)+1}{1-t} dt = \int_0^y (-1 + \frac{1}{1-t}) dt = [-t - \log(1-t)]_0^y = -y - \log(1-y)$. So, $\sum_{n=1}^{\infty} \frac{y^{n+1}}{n+1} = -y - \log(1-y)$. Let $m = n+1$. Then $\sum_{m=2}^{\infty} \frac{y^m}{m} = -y - \log(1-y)$. We know $\sum_{m=1}^{\infty} \frac{y^m}{m} = -\log(1-y)$. So, $(-\log(1-y)) - y = -y - \log(1-y)$, which is consistent.

We have $I = \sum_{n=1}^{\infty} \frac{1}{n(n+1)7^n}$. Let's use the identity: $\frac{1}{n(n+1)} = \int_0^1 t^{n-1}(1-t) dt$. $I = \sum_{n=1}^{\infty} \left(\int_0^1 t^{n-1}(1-t) dt\right) \frac{1}{7^n}$ $I = \int_0^1 (1-t) \sum_{n=1}^{\infty} \frac{t^{n-1}}{7^n} dt$ $I = \int_0^1 (1-t) \frac{1}{7} \sum_{n=1}^{\infty} \frac{t^{n-1}}{7^{n-1}} dt$ Let $m = n-1$. When $n=1$, $m=0$. $I = \int_0^1 (1-t) \frac{1}{7} \sum_{m=0}^{\infty} \left(\frac{t}{7}\right)^m dt$ The sum is a geometric series: $\sum_{m=0}^{\infty} r^m = \frac{1}{1-r}$ for $|r|<1$. Here $r = t/7$. Since $0 \le t \le 1$, $|t/7| \le 1/7 < 1$. So the sum is $\frac{1}{1 - t/7} = \frac{7}{7-t}$. $I = \int_0^1 (1-t) \frac{1}{7} \cdot \frac{7}{7-t} dt$ $I = \int_0^1 \frac{1-t}{7-t} dt$

Now, we evaluate this integral. Let $u = 7-t$. Then $t = 7-u$, and $dt = -du$. When $t=0$, $u=7$. When $t=1$, $u=6$. $I = \int_{7}^{6} \frac{1-(7-u)}{u} (-du)$ $I = \int_{7}^{6} \frac{1-7+u}{u} (-du)$ $I = \int_{7}^{6} \frac{u-6}{u} (-du)$ $I = \int_{6}^{7} \frac{u-6}{u} du$ $I = \int_{6}^{7} \left(1 - \frac{6}{u}\right) du$ $I = \left[u - 6 \log_e|u|\right]_{6}^{7}$ $I = (7 - 6 \log_e 7) - (6 - 6 \log_e 6)$ $I = 7 - 6 \log_e 7 - 6 + 6 \log_e 6$ $I = 1 + 6 (\log_e 6 - \log_e 7)$ $I = 1 + 6 \log_e \left(\frac{6}{7}\right)$

This matches option (A).

Let's double-check the steps and reasoning. The substitution $y=1/x$ was correct. The splitting of the integral $\int_{1}^{\infty} \frac{1}{y^2 7^{[y]}} dy$ into $\sum_{n=1}^{\infty} \int_{n}^{n+1} \frac{1}{y^2 7^n} dy$ was correct. The integration of $1/y^2$ was correct. The resulting series was $\sum_{n=1}^{\infty} \frac{1}{n(n+1)7^n}$.

The identity $\frac{1}{n(n+1)} = \int_0^1 t^{n-1}(1-t) dt$ is correct. We can verify this: $\int_0^1 t^{n-1}(1-t) dt = \int_0^1 (t^{n-1} - t^n) dt = \left[\frac{t^n}{n} - \frac{t^{n+1}}{n+1}\right]_0^1 = \frac{1}{n} - \frac{1}{n+1} = \frac{n+1-n}{n(n+1)} = \frac{1}{n(n+1)}$.

The interchange of integral and summation is justified because the series converges uniformly on $[0, 1]$ for the variable $t$.

The evaluation of the integral $\int_0^1 \frac{1-t}{7-t} dt$ was correct. The substitution $u=7-t$ was correct. The final calculation of the definite integral was correct.

Common Mistakes & Tips

• Handling the Greatest Integer Function: Be careful when splitting the integral. The intervals are typically of the form $[n, n+1)$ or $(n, n+1]$. For $[1/x]$, the intervals are of the form $(1/(n+1), 1/n]$.
• Interchange of Summation and Integration: Ensure that the conditions for interchanging summation and integration are met. In this case, the uniform convergence of the series $\sum_{n=1}^{\infty} \frac{t^{n-1}}{7^n}$ on the interval $[0, 1]$ justifies this step.
• Logarithm Properties: Remember that $\log_e(a/b) = \log_e a - \log_e b$. This was used in simplifying the final expression.

Summary

The integral was evaluated by first performing a substitution to simplify the argument of the greatest integer function. The resulting integral was then split into an infinite series of integrals over unit intervals. The integrand in each interval was constant with respect to the integration variable, allowing for straightforward integration. The resulting series was then manipulated using a known integral representation of the term $1/(n(n+1))$, which allowed for the interchange of summation and integration. The final integral was evaluated using a further substitution, leading to the result in terms of logarithms.

The final answer is $1 + 6 \log_e(6/7)$.

The final answer is \boxed{1 + 6{\log _e}\left( {{6 \over 7}} \right)}.