integration - Proving a generalisation of the integral $int_0^inftyfrac{sin(x)}{x}dx$

Plugging generalisations of the integral $\int_0^\infty\frac{\sin(x)}{x}dx$ into Wolfram Alpha has lead me to conjecturing the following result:

\begin{align}
\int_0^\infty\frac{\sin(x^n)}{x}dx=\frac{\pi}{2n} \tag1
\end{align}

where $n$ is an arbitrary natural number. It holds for values of $n$ from $1$ to $10$, as I've checked using Wolfram Alpha.

However, I haven't had any success computing the integral directly to prove this result. I've seen the case for $n=1$ evaluated using differentiation under the integral sign, so I tried to do that.

We define

\begin{align}
I(n,b):=\int_0^\infty\frac{\sin(x^n)}{x}e^{-bx}dx = \Im\int_0^\infty\frac{e^{ix^n-bx}}{x}dx \tag2
\end{align}

Differentiating $(2)$ with respect to $b$, we get

\begin{align}
\frac{\partial I(n,b)}{\partial b} = -\Im\int_0^\infty e^{ix^n-bx}dx \tag3

\end{align}

This takes care of the denominator and the negative real part of exponent ensures that the integral converges, but I have no idea how to compute it.

Another approach I tried is to define

\begin{align}
I(n):=\int_0^\infty\frac{\sin(x^n)}{x}dx \tag4
\end{align}

and then to differentiate this with respect to $n$, yielding

\begin{align}
I'(n) = \int_0^\infty\log(x)x^{n-1}\sin(x^n)dx \tag5
\end{align}

which, using the substitution $z=x^n$, $dz = nx^{n-1}dx$, becomes

\begin{align}
I'(n)&=\frac{1}{n}\int_0^\infty\log(\sqrt[n]{z})\sin(z)dz\\

&= \frac{1}{n^2}\int_0^\infty\log(z)\sin(z)dz \tag6
\end{align}

However, $(6)$ is divergent, so it's not useful in computing $I(n)$ either.

Does anybody have any insight into how to make differentiation under the integral sign work for this problem or how to calculate $(3)$, or is it that wrong approach entirely? If it's the wrong approach, how would one go about proving (or disproving, if the whole thing is wrong) my conjectured result?