Convergence of integral $\int_{17}^\infty \frac{(\ln{\ln{x})}\sin{(e^{ax}})}{x+e}$ For which $a\in\mathbb{R}$ does the integral $\int_{17}^\infty \frac{(\ln{\ln{x})}\sin{(e^{ax}})}{x+e}$ converge?
I tried with substitution $t=e^{ax}$, but don't know how to do the rest?
Any help is welcome. Thanks in advance
 A: If $\;a<0,\;$ then $\;\sin(e^{ax})<e^{ax},\;$ and
$$I<\int\limits_{17}^\infty e^{ax}\,\text dx,$$
i.e. converges.
If $\;a=0,\;$ then
$$I \ge \dfrac{17\sin1}{17+e} \int\limits_{17}^\infty \dfrac{\ln\ln x}xdx
= \dfrac{17\sin1}{17+e} \int\limits_{17}^\infty \ln\ln x\,\text d\ln x,$$
i.e. diverges.
If $\;a>0,\; x\ge 17 > e^e,\;$ then
$$\;\dfrac{d}{dx}\dfrac{\ln \ln x}{x+e} = \dfrac{e-x(\ln x\cdot\ln\ln x-1)}{x(x+e)^2\ln x} < 0.\tag1$$
Applying substitution $\;x=\dfrac1a\ln y,\; k_0=\left\lceil\dfrac{e^{17a}}{2\pi}\right\rceil,\;$ easily to get
$$I = \dfrac1a\int\limits_{\large e^{17a}}^\infty\;  \dfrac{\ln\ln\left(\dfrac1a\ln y\right)}{\dfrac1a\ln y+e}\,\dfrac{\sin y}y\,\text dy,$$
$$aI = \int\limits_{e^{17a}}^{2k_0\pi}\; \dfrac{\ln\ln\left(\dfrac1a\ln y\right)}{\dfrac1a\ln y+e}\,\dfrac{\sin y}y\,\text dy
+ \sum\limits_{k=k_0}^\infty\;\int\limits_{2\pi k}^{2\pi k+2\pi}\; \dfrac{\ln\ln\left(\dfrac1a\ln y\right)}{\dfrac1a\ln y+e}\,\dfrac{\sin y}y\,\text dy.\tag2 $$
From $(1)$ should, that the sinusoids in the integrals under the sum have decreasing weight, and for $\;y\in(2k\pi,2k\pi+\pi),\;k=k_0\dots\infty\;$
$$\dfrac{\ln\ln\left(\dfrac1a\ln y\right)}{\dfrac1a\ln y+e}\,\dfrac{\sin y}y
\ge\left|\dfrac{\ln\ln\left(\dfrac1a\ln(y+\pi)\right)}{\dfrac1a\ln(y+\pi)+e}\,\dfrac{\sin (y+\pi)}{y+\pi}\right|.$$
Then
$$\int\limits_{2\pi k}^{2\pi k+2\pi}\; \dfrac{\ln\ln\left(\dfrac1a\ln y\right)}{\dfrac1a\ln y+e}\,\dfrac{\sin y}y\,\text dy>0.$$
Therefore, should exist the constants $\;C_1,\,C_2,\;$ such as
$$I < C_1 + C_2\int\limits_{2\pi k_0}^\infty\dfrac{\sin y}y\,\text dy,\tag3$$
i.e. converges.
Finally, the given integral converges if $\;a\in\mathbb R\setminus \{0\}.$
