Convoluted Integral

$\large \displaystyle I = \int_1^{\infty} \frac{\ln(\ln(x))}{x^{2019}} dx$

Let $x=e^u$ . Then $dx=e^u du$ and:

$\large \displaystyle I = \int_0^{\infty} e^{-2018u}\ln(u)du$

Let $u = \frac{t}{2018}$ . Then $du = \frac{dt}{2018}$ and:

$\large \displaystyle I = \frac{1}{2018} \left ( \int_0^{\infty} e^{-t}\ln \left ( \frac{t} {2018} \right) dt \right )$

$\large \displaystyle I = \frac{1}{2018} \left ( \int_0^{\infty} e^{-t}\ln(t)dt - \int_0^{\infty} e^{-t}\ln(2018)dt \right)$

$\large \displaystyle I = - \frac{1}{2018} \left ( - \int_0^{\infty} e^{-t}\ln(t)dt + \ln(2018) \int_0^{\infty} e^{-t}dt \right)$

The first integral is the definition of the Euler-Mascheroni constant , $\gamma$ :

Then :

$\color{#20A900} \boxed{\large \displaystyle I = - \frac{\gamma + \ln(2018)}{2018}}$

Thus:

$\color{#3D99F6} \large \displaystyle A=2018, B=1, C=2018, \boxed{\large \displaystyle A+B+C=4037}$

Hey there @Hamza A , I believe it was Hummus A previously at least 4 years ago when I was active on this platform. We know from the definition of Gamma function that,

$\displaystyle \Gamma(a) = 2018^a \int_0^{\infty} e^{-2018x} x^{a-1} \operatorname{dx}$ Differentiating both sides with respect to 'a' we have, $\displaystyle \dfrac{\Gamma(a)}{2018^a}\left(\psi_{0}(a)-\ln(2018)\right) = \int_0^{\infty} e^{-2018x} x^{a-1} \ln x \operatorname{dx}$ Putting $a=1$ along with $\psi_0(1)=-\gamma$ gives us, $\displaystyle \int_0^{\infty} e^{-2018x} \ln x \operatorname{dx} = -\dfrac{\ln(2018)+\gamma}{2018}$ Substitute $e^x \to x$ to get, $\displaystyle \int_1^\infty \dfrac{\ln(\ln x)}{x^{2019}} \operatorname{dx} =\int_0^{\infty} e^{-2018x} \ln x \operatorname{dx} = -\dfrac{\ln(2018)+\gamma}{2018}$ Therefore the answer is, $\boxed{A+B+C = 2018+1+2018 = 4037}$

Slightly different solution, but using differentiation and the Gamma function:

$\displaystyle I=\int_1^{\infty}\frac{\ln(\ln(x))}{x^{2019}}dx$

Substitute $x=e^u,\quad dx=e^udu$

$\displaystyle I=\int_0^{\infty}\frac{\ln(u)}{e^{2018u}}du = \frac{d}{dn} \int_0^{\infty}\frac{u^{n-1}}{e^{2018u}}du |_{n=1}$

Substitute $t=2018u$

$\displaystyle I=\frac{d}{dn}\frac{1}{2018^{n}}\int_0^{\infty}\frac{t^{n-1}}{e^{t}}dt |_{n=1}$

$\displaystyle I=\frac{d}{dn}\frac{\Gamma(n)}{2018^n}|_{n=1}$

$\displaystyle I= -\frac{\Gamma(n)(\ln{2018} - ψ^{(0)}(n)))}{2018^n}|_{n=1} = \boxed{-\frac{\ln{2018}-\gamma}{2018}}$