Solving $\int_0^{\infty} \ln^m(x)\sin\left(x^n\right)\:dx$

Using your own parameterized integral $J_n(\phi)$ we may use Ramanujan's Master Theorem.

Ramanujan's Master Theorem $($RMT$)$

Let $f(x)$ be an analytic function with a MacLaurin Expansion of the form $$f(x)=\sum_{k=0}^{\infty}\frac{\varphi(k)}{k!}(-x)^k$$then the Mellin Transform of this function is given by $$\int_0^{\infty}x^{s-1}f(x)dx=\Gamma(s)\varphi(-s)$$

For the purpose of exploiting this theorem we expand the sine function as its corresponding MacLaurin Series and enforce the substitution $x^n\mapsto x$ firstly and $x^2\mapsto x$ afterwards $($Note: the distinction between $x^n\mapsto x$ and $x^2\mapsto x$ instead of directly $x^{2n}\mapsto x$ is only chosen for simplicity$)$. This leads to

\begin{align*} J_n(\phi)=\int_0^\infty x^{\phi}\sin(x^n)\mathrm dx &= \int_0^\infty x^{\phi}\left[\sum_{k=0}^\infty(-1)^k\frac{(x^n)^{2k+1}}{(2k+1)!}\right]\mathrm dx\\ &=\int_0^\infty x^{\phi/n}\left[\sum_{k=0}^\infty(-1)^k\frac{x^{2k+1}}{(2k+1)!}\right]\frac{\mathrm dx}{nx^{1-1/n}}\\ &=\frac1n\int_0^\infty x^{(\phi+1)/n}\left[\sum_{k=0}^\infty(-1)^k\frac{x^{2k}}{(2k+1)!}\right]\mathrm dx\\ &=\frac1n\int_0^\infty x^{(\phi+1)/2n}\left[\sum_{k=0}^\infty(-1)^k\frac{x^k}{(2k+1)!}\right]\frac{\mathrm dx}{2x^{1/2}}\\ &=\frac1{2n}\int_0^\infty x^{(\phi+1)/2n-1/2}\sum_{k=0}^\infty\frac{\Gamma(k+1)/\Gamma(2k+2)}{k!}(-x)^k\mathrm dx \end{align*}

The new gained structure is cleary recognizable as possible application of the RMT. Therefore set $s=\frac{\phi+1}{2n}+\frac12$ and moreover $\varphi(k)=\frac{\Gamma(k+1)}{\Gamma(2k+2)}$. Thus, we get

\begin{align*} J_n(\phi)&=\frac1{2n}\int_0^\infty x^{(\phi+1)/2n-1/2}\sum_{k=0}^\infty\frac{\Gamma(k+1)/\Gamma(2k+2)}{k!}(-x)^k\mathrm dx\\ &=\frac1{2n}\Gamma\left(\frac{\phi+1}{2n}+\frac12\right)\frac{\Gamma\left(1-\left(\frac{\phi+1}{2n}+\frac12\right)\right)}{\Gamma\left(2-2\left(\frac{\phi+1}{2n}+\frac12\right)\right)}\\ &=\frac1{2n}\frac{\Gamma\left(\frac12+\frac{\phi+1}{2n}\right)\Gamma\left(\frac12-\frac{\phi+1}{2n}\right)}{\Gamma\left(1-\frac{\phi+1}n\right)}\\ &=\frac1{2n}\frac1{\Gamma\left(1-\frac{\phi+1}n\right)}\frac{\pi}{\sin\left(\frac{\phi+1}{2n}\pi+\frac\pi2\right)}\\ &=\frac1{n}\frac1{\Gamma\left(1-\frac{\phi+1}n\right)}\frac{\pi}{2\cos\left(\frac{\phi+1}{2n}\pi\right)}\frac{\sin\left(\frac{\phi+1}{2n}\pi\right)}{\sin\left(\frac{\phi+1}{2n}\pi\right)}\\ &=\frac1n\Gamma\left(\frac{\phi+1}n\right)\sin\left(\frac{\phi+1}{2n}\pi\right) \end{align*}

$$\therefore~J_n(\phi)~=~\int_0^\infty x^{\phi}\sin(x^n)\mathrm dx~=~\frac1n\Gamma\left(\frac{\phi+1}n\right)\sin\left(\frac{\phi+1}{2n}\pi\right)$$

From hereon we can deduce the same formulae ComplexYetTrivial's answer contains, and which he already did quite well by invoking Leibniz rule and Faà di Bruno's formula. The crucial point of this post is to present another possible derivation in order to obtain the Mellin Transform of the sine function. In my opionion using the RMT explains the close connection to the Gamma Function quite well. Of course, one has to be careful with the choice of $\phi$ and $n$ hence an occuring negative integer value within on of the Gamma Functions in the nominator would cause an indefinite expression.

$$\therefore~I_{m,n}~=~\int_0^\infty \ln^m(x)\sin(x^n)\mathrm dx~=~\lim_{\phi\to0}\frac{\mathrm d^m}{\mathrm d\phi^m}\left[\frac1n\Gamma\left(\frac{\phi+1}n\right)\sin\left(\frac{\phi+1}{2n}\pi\right)\right]$$

A method relying on the Mellin transform of the sine:

For $s>1$ and $z \in \mathbb{C}$ with $-s < \operatorname{Re} (z) < s$ we have $$ f_s (z) \equiv \int \limits_0^\infty x^{z-1} \sin(x^s) \, \mathrm{d} x = \frac{1}{s} \int \limits_0^\infty t^{\frac{z}{s}-1} \sin(t) \, \mathrm{d} t = \frac{1}{s} \mathcal{M}(\sin) \left(\frac{z}{s}\right) = \frac{1}{s} \sin \left(\frac{\pi z}{2s}\right) \operatorname{\Gamma} \left(\frac{z}{s}\right) $$ with the limit $f_s(0) = \frac{\pi}{2s}$. For $m \in \mathbb{N_0}$ and $s>1$ this implies \begin{align} I_{m,s} &\equiv \int \limits_0^\infty \ln^m(x) \sin(x^s) \, \mathrm{d} x = f_s^{(m)}(1) = \frac{1}{s} \frac{\mathrm{d}^m}{\mathrm{d} z^m} \left[\sin \left(\frac{\pi z}{2s}\right) \operatorname{\Gamma} \left(\frac{z}{s}\right)\right] \Bigg\vert_{z=1} \\ &= \frac{1}{s^{m+1}} \frac{\mathrm{d}^m}{\mathrm{d} x^m} \left[\sin \left(\frac{\pi }{2}x\right) \operatorname{\Gamma} \left(x\right)\right] \Bigg\vert_{x= 1/s} = \frac{1}{s^{m+1}} \sum \limits_{k=0}^m {m \choose k} \left(\frac{\pi}{2}\right)^k \sin^{(k)} \left(\frac{\pi}{2s}\right) \operatorname{\Gamma}^{(m-k)} \left(\frac{1}{s}\right) \, , \end{align} where the last step follows from the general Leibniz rule. The derivatives of the gamma function can be expressed in terms of polygamma functions using Faà di Bruno's formula, but otherwise that is probably as elementary as it gets.

In the special case $s=2$ we can use $$ \sin^{(k)} \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} (-1)^{\lfloor k/2 \rfloor} $$ for $k \in \mathbb{N}_0$ and the values (obtained from the Legendre duplication formula) \begin{align} \operatorname{\Gamma} \left(\frac{1}{2}\right) &= \sqrt{\pi} \, , \\ \operatorname{\psi}^{(0)} \left(\frac{1}{2}\right) &= - \gamma - 2 \ln(2) \, , \\ \operatorname{\psi}^{(n)} \left(\frac{1}{2}\right) &= (-1)^{n-1} n! (2^{n+1}-1) \zeta(n+1) \, , \, n \in \mathbb{N} \, , \end{align} to simplify the final result. The complexity of Faà di Bruno's formula prevents us from finding a reasonably nice general expression for $(I_{m,2})_{m \in \mathbb{N}_0}$ , but at least we know that these integrals can be written in terms of $\pi$, $\ln(2)$, $\gamma$ and zeta values.