Easier than it looks

Calculus Level 5

$\left(\frac{\Im \left(\displaystyle\int_{0}^{\frac{1}{e}}\ln{(1+\ln{x})}\ \mathrm dx\right )}{\Re\left (\displaystyle\int_{0}^{\frac{1}{e}}\ln{(1+\ln{x})}\ \mathrm dx\right )}\right)^2 = \ ?$

Notations: $\Im(\cdot)$ and $\Re(\cdot)$ respectively denote the imaginary and real parts of a complex number .

The answer is 29.6226241172.

1 solution

คลุง แจ็ค
Jan 20, 2020

We first substitute $\displaystyle x=e^{u}$ , so, we get

$\displaystyle \int_{-\infty}^{-1} e^{u} \ln(1+u)du$

Next, we substitute $\displaystyle v=u+1$

$\displaystyle \int_{-\infty}^{0} e^{v-1} \ln(v)dv$

Lastly, set $\displaystyle y=-v$

$\displaystyle \int_{0}^{\infty} e^{-y-1} \left (i \pi+\ln(y) \right )dy=\dfrac{1}{e} \left ( i \pi \Gamma(1)+\Gamma'(1) \right )=\dfrac{1}{e} \left ( i \pi-\gamma \right )$

So, the answer is $\displaystyle \boxed{\left ( \dfrac{\pi}{\gamma} \right)^{2} \approx 29.6226241172}$

Easier than it looks

The answer is 29.6226241172.

1 solution

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