The Logarithmic Reciprocal

Let us first calculate an upper bound for the integral $\displaystyle \int_2^e \! \frac {1}{ln(x)} \, \mathrm{d}x$ . The graph of $f(x)= \frac {1}{ln(x)}$ is already provided in the question as reference. Note that f(x) decreases from $f(2)= \frac {1}{ln(2)} \approx 1.443$ to $f(e)=1$ in the region $[2,e]$ . Since the value of $\displaystyle \int_2^e \! \frac {1}{ln(x)} \, \mathrm{d}x$ is the area under the curve $y=f(x)= \frac {1}{ln(x)}$ in the region $[2,e]$ , it must be intuitively less than the area of the trapezoid shown in the figure below.

Therefore, $\displaystyle \int_2^e \! \frac {1}{ln(x)} \, \mathrm{d}x\;<\; \frac {1}{2} (1.443+1)(0.718) \approx 0.877$

Next, we move to the more involved task of calculating a lower bound for the integral $\displaystyle \int_2^e \! \frac {1}{ln(x)} \, \mathrm{d}x$ . Using the definition of definite integral as a limit, we get

$\displaystyle \int_a^b \! f(x) \, \mathrm{d}x$ $= \displaystyle \lim_{n \to \infty} \displaystyle \sum_{i=1}^{n} ( \frac {b-a}{n}) f(a+ \frac {b-a}{n}i)$

and,

$\displaystyle \int_a^b \! \frac {1}{f(x)} \, \mathrm{d}x$ $= \displaystyle \lim_{n \to \infty} \displaystyle \sum_{i=1}^{n} ( \frac {b-a}{n}) \frac {1}{f(a+ \frac {b-a}{n}i)}$

Now, if $f(x)$ takes only positive values in the region $[a,b]$ , then we can use the AM-HM Inequality to write

$\frac { \displaystyle \sum_{i=1}^{n} f(a+ \frac {b-a}{n}i)}{ \displaystyle n} \geq \frac { \displaystyle n}{ \displaystyle \sum_{i=1}^{n} \frac {1}{f(a+ \frac {b-a}{n}i)}}$

$\Rightarrow ( \displaystyle \frac {1}{n} \displaystyle \sum_{i=1}^{n} f(a+ \frac {b-a}{n}i))( \displaystyle \frac {1}{n} \displaystyle \sum_{i=1}^{n} \frac {1}{f(a+ \frac {b-a}{n}i)}) \geq 1$

$\Rightarrow ( \displaystyle \frac {b-a}{n} \displaystyle \sum_{i=1}^{n} f(a+ \frac {b-a}{n}i))( \displaystyle \frac {b-a}{n} \displaystyle \sum_{i=1}^{n} \frac {1}{f(a+ \frac {b-a}{n}i)}) \geq (b-a)^{2}$

Therefore,

$( \displaystyle \lim_{n \to \infty} \displaystyle \sum_{i=1}^{n} ( \frac {b-a}{n}) f(a+ \frac {b-a}{n}i))$

$\;\;\;\times (\displaystyle \lim_{n \to \infty} \displaystyle \sum_{i=1}^{n} ( \frac {b-a}{n}) \frac {1}{f(a+ \frac {b-a}{n}i)}) \geq (b-a)^{2}$

$\Rightarrow ( \displaystyle \int_a^b \! f(x) \, \mathrm{d}x) \times ( \displaystyle \int_a^b \! \frac {1}{f(x)} \, \mathrm{d}x) \geq (b-a)^{2}$

Substituting $f(x)=ln(x)$ , $a=2$ and $b=e$ , we get

$( \displaystyle \int_2^e \! ln(x) \, \mathrm{d}x) \times ( \displaystyle \int_2^e \! \frac {1}{ln(x)} \, \mathrm{d}x) \geq (e-2)^2 \approx 0.5155$

Now, it is easy to calculate that $\displaystyle \int_2^e \! ln(x) \, \mathrm{d}x = 2-2ln(2) \approx 0.614$

Therefore, $\displaystyle \int_2^e \! \frac {1}{ln(x)} \, \mathrm{d}x \geq \frac {0.5155}{ \displaystyle \int_2^e \! ln(x) \, \mathrm{d}x} = \frac {0.5155}{0.614} \approx 0.8396$

After all these calculations we get $0.8396 \leq \displaystyle \int_2^e \! \frac {1}{ln(x)} \, \mathrm{d}x < 0.877$

$\Rightarrow 8.396 \leq 10 \times \displaystyle \int_2^e \! \frac {1}{ln(x)} \, \mathrm{d}x < 8.77$

Therefore, $[\;10 \times \displaystyle \int_2^e \! \frac {1}{ln(x)} \, \mathrm{d}x\;] = 8$

NOTE:

For any function $f(x)$ , if $M$ and $m$ are respectively the global maximum and global minimum of $f(x)$ in the region $[a,b]$ , then

$m(b-a) \leq \displaystyle \int_a^b \! f(x) \, \mathrm{d}x \leq M(b-a)$

In this case, $f(x)= \frac {1}{ln(x)}$ , $a=2$ and $b=e$ . It might be tempting to use the above inequality for this problem. Note that $m=1$ and $M \approx1.443$ . So, we have

$0.718 \leq \displaystyle \int_2^e \! \frac {1}{ln(x)} \, \mathrm{d}x \leq 1.036$

This inequality is indeed true, but we need a much tighter bound to get the exact value of $[\;10 \times \displaystyle \int_2^e \! \frac {1}{ln(x)} \, \mathrm{d}x\;]$ . The other 3 incorrect options in the question ( $7$ , $9$ and $10$ ) are mentioned specifically keeping this in mind!

Please note that I have developed the idea & then designed the question myself. The lower bound could also have been derived using the Cauchy-Schwartz Inequality, but I found using the AM-HM Inequality very exciting! In case there are other possible solutions, please mention in the comments.

Disclaimer: I won't be going through my motivations, or it will be a really long solution. If you would like to know why I did what I did, comment and I will get back to you.

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$y = \frac{1}{\ln x}$ passes through $(2, \frac{1}{\ln 2})$ and $(e, 1)$ . The secant is of gradient $\displaystyle \frac{1-\frac{1}{\ln 2}}{e-2} \approx -0.617$ . By MVT, there exists some point with tangent parallel to the secant.

$\frac{d}{dx} \frac{1}{\ln x} = -0.617 \longrightarrow x\ln^{2} x = 1.621$

For $x = 2$ , $x\ln^{2} x = 2(0.693)^{2} \approx 0.960$ . For $x = e$ , $x\ln^{2} x = e(1)^{2} = 2.178$ .

Since $x\ln^{2} x$ is monotonically increasing, we choose a convenient $x$ "halfway" between $2$ and $e$ : the geometric mean. Verifying, we see $x\ln^{2} x = \sqrt{2e}(\frac{\ln 2 + 1}{2})^{2} \approx 1.671$ .

So $\frac{1}{\ln \sqrt{2e}} = \frac{2}{\ln 2 + 1} \approx 1.181$ and $\displaystyle \frac{\frac{1}{\ln 2}+1}{2} \approx 1.222$ . Hence, the midpoint is $\frac{1.181+1.222}{2} = 1.2015$ . The line that passes through $(\sqrt{2e}, 1.2015)$ with gradient $-0.617$ is $y = -0.617x+2.640$ .

So the area is approximately

$\int_{2}^{2.718} -0.617x+2.640 \ dx \approx \boxed{0.851}$

Quite accurate, considering WolframAlpha says it's $0.8500$ .

I personally would use the trapezium rule with 8 strips from 2.0 to 2.7: $\ Area \approx \frac{h}{2}( y_0 +y_n + 2 \displaystyle\sum_{k=1}^{n-1} y_k)$ $\Rightarrow\ Area \approx 0.05( \frac{1}{ln2} + \frac{1}{Ln 2.7} + 2(\frac{1}{ln2.1} +...+\frac{1}{Ln2.6} ) )$ $= 8.32 \ (3s.f.)$ From the graph it is clear that because the gradient is so shallow so the overestimate will be very small. $\therefore\lfloor 10 \times \int_{2}^{e} \frac{1}{ln(x)} dx \rfloor = \boxed{8}$

I calculate the upper bound the same way you did.

For the lower bound I make the substitution $x=e^{z}$ So the integral becomes $\int_1^{ln2} \frac{e^z}{z} dz$

Now in the region $[ \ln 2,1] \frac{e^z}{z}$ is decreasing function.

So $\frac{e}{1} \int_{ln2}^1 dz < \int_{ln2}^1 \frac{e^z}{z} dz < \frac{e^2}{ln2} \int_{ln2}^1 dz$

$\Longrightarrow \int_{ln2}^1 \frac{e^z}{z} dz >e \int_{ln2}^1 dz =e (1-ln2)=0.834$

I've used just a basic form of definite integral to make this solve..