## May 21, 2018

Have deleted answer due to an error.

May 21, 2018

$\textcolor{b l u e}{{\int}_{a}^{2 a} \frac{1}{x} \cdot \mathrm{dx} = {\int}_{3 a}^{6 a} \frac{1}{x} \cdot \mathrm{dx} = 0.693}$

#### Explanation:

The first integral

${\int}_{a}^{2 a} \frac{1}{x} \cdot \mathrm{dx} = {\left[\ln x\right]}_{a}^{2 a} = \ln 2 a - \ln a = \ln 2 + \ln a - \ln a = \ln 2 = 0.693$

The second integral

${\int}_{3 a}^{6 a} \frac{1}{x} \cdot \mathrm{dx} = {\left[\ln x\right]}_{3 a}^{6 a} = \ln 6 a - \ln 3 a$

$= \ln 6 + \ln a - \ln 3 - \ln a = \ln 6 - \ln 3 = 1.792 - 1.099 = 0.693$

Note that the two integrals have same value

$\textcolor{b l u e}{{\int}_{a}^{2 a} \frac{1}{x} \cdot \mathrm{dx} = {\int}_{3 a}^{6 a} \frac{1}{x} \cdot \mathrm{dx} = 0.693}$

note that

$\textcolor{red}{\ln \left(x \cdot y\right) = \ln x + \ln y}$

May 21, 2018

The two integrals are identical.

#### Explanation:

We seek the greater of:

${I}_{1} = {\int}_{a}^{2 a} \setminus \frac{1}{x} \setminus \mathrm{dx}$ or ${I}_{2} = {\int}_{3 a}^{6 a} \setminus \frac{1}{x} \setminus \mathrm{dx}$

Where $a > 0$. Consider the general case:

$I \left(\alpha , \beta\right) = {\int}_{\alpha}^{\beta} \setminus \frac{1}{x} \setminus \mathrm{dx}$

Where $\alpha , \beta > 0$, Which is a standard integral, so we can directly integrate:

$I \left(\alpha , \beta\right) = {\left[\setminus \ln | x | \setminus\right]}_{\alpha}^{\beta}$

And as $\alpha , \beta > 0$, we can additionally use the properties of logarithms to write:

$I \left(\alpha , \beta\right) = \ln \beta - \ln \alpha = \ln \left(\frac{\beta}{\alpha}\right)$

This integral is in fact used to define the Napier logarithm and the unexpected ratio is an alternative proof of the logarithm of a product property! Given this result we now conclude that

${I}_{1} = \ln \left(\frac{2 a}{a}\right) = \ln 2$
${I}_{2} = \ln \left(\frac{6 a}{3 a}\right) = \ln 2$

Making the two integrals identical