Taking a Dip II

$\begin{aligned} I(n) & = \int_1^\infty \frac {\ln(\ln x)}{x^{n+1}}dx & \small \color{#3D99F6} \text{Let }u = \ln x \implies e^u = x \implies e^u du = dx \\ & = \int_0^\infty \frac {\ln u}{e^{nu}}du & \small \color{#3D99F6} \text{By differentiation under an integral sign} \\ \frac {\partial I(n)}{\partial n} & = - \int_0^\infty u e^{-nu} \ln u \ du & \small \color{#3D99F6} \text{By integration by parts} \\ & = \frac {u e^{-nu} \ln u }n \bigg|_0^\infty - \frac 1n \int_0^\infty e^{-nu}(\ln u + 1) \ du \\ & = {\color{#3D99F6} \lim_{u \to \infty} \frac {u \ln u}{ne^{nu}} - \lim_{u \to 0} \frac {\ln u}{\frac {ne^{nu}}u}} - \frac {I(n)}n+ \frac {e^{-nu}}{n^2}\bigg|_0^\infty & \small \color{#3D99F6} \text{By L'Hôpital's rule} \\ & = {\color{#3D99F6} \lim_{u \to \infty} \frac {\ln u+1}{n^2e^{nu}} - \lim_{u \to 0} \frac {\frac 1u}{\frac {n^2ue^{nu}-ne^{nu}}{u^2}}} - \frac 1{n^2} \\ & = \lim_{u \to \infty} \frac {\frac 1u}{n^3e^{nu}} - \lim_{u \to 0} \frac u{n^2ue^{nu}-ne^{nu}} - \frac {I(n)}n - \frac 1{n^2} \\ & = 0 - 0 - \frac {I(n)}n - \frac 1{n^2} \end{aligned}$

$\begin{aligned} \implies I(n) + n \frac {\partial I(n)}{\partial n} & = - \frac 1n \\ \frac {\partial nI(n)}{\partial n} & = - \frac 1n \\ \implies n I(n) & = {\color{#3D99F6} C} - \ln n & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \\ (1) (- \gamma) & = C - \ln 1 & \small \color{#3D99F6} \text{Note that }I(1) = \int_0^\infty e^{-u} \ln u \ du = - \gamma \\ \implies I(n) & = - \frac {\gamma + \ln n}n & \small \color{#3D99F6} \text{Note that }C = - \gamma \end{aligned}$

To find the minimum value of $I(n)$ ,

$\begin{aligned} \frac {d I(n)}{dn} & = - \frac {1-(\gamma + \ln n)}{n^2} & \small \color{#3D99F6} \text{Equating }\frac {dI(n)}{dn} = 0 \\ \implies \ln n & = 1-\gamma \\ n & = e^{1-\gamma} & \small \color{#3D99F6} \text{Since }\frac {d^2I(n)}{dn^2} > 0 \text{ when } n = e^{1-\gamma} \text{ (see note)} \\ \min (I(n)) & = I(e^{1-\gamma}) \\ & = - \frac {\gamma + 1 -\gamma}{e^{1-\gamma}} \\ & = - e^{\gamma -1} \end{aligned}$

$\implies A+B = -1-1 = \boxed{-2}$

$$

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Note:

$\begin{aligned} \frac {d^2I(n)}{dn^2} & = \frac {1-(\ln n + \gamma -1)(2n)}{n^3} & \small \color{#3D99F6} \text{Putting }n=e^{1-\gamma} \\ I''(e^{1-\gamma}) & = \frac {1-0}{e^{3(1-\gamma)}} > 0 \end{aligned}$

Let $I$ be the integral shown above. We consider the substitution $u=\ln x$ so that the integral becomes $I=\int^{\infty}_0e^{-nu}\ln(u)\ du.$ Observe that this is simply the Laplace transform of the natural logarithm function, so $I=-\frac{1}{n}(\ln(n)+\gamma).$ The proof can be found here . Now, it remains to set the derivative of $I$ to 0 to obtain that the minimum is reached when $n=e^{1-\gamma}.$ Substituting back the value of $n$ yields the minimum value of $I=-e^{1-\gamma}$ and thus $A=-1, B=-1\Rightarrow A+B=-2.$