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?
