I am trying to prove the following formula for $\operatorname{Re} z>0$
$$\frac{\Gamma(z)\sin\theta z}{n(a^2+b^2)^{z/2}} = \int_0^\infty e^{-at^n}t^{nz-1} \sin(bt^n)\,dt$$
where $n$ is a positive integer, $a$ and $b$ are real numbers with $(a, b) \neq (0,0)$, and $\tan = b/a$. When $a=0$, we take $\theta = \pm \pi/2$ such that $b>0$.
I used $\Gamma(z)= \int_0^\infty e^{-s}s^{z-1}\,ds$ with the substitution $s=(a+ib)t^n$ based on the given hint, but couldn't prove it. Any help is appreciated.
First we notice that the given integral is equal to this:
$$I=\int_0^\infty e^{-at^n}t^{nz-1} \sin(bt^n)dt=\frac{1}{2i}\left[\int_0^\infty e^{-at^n}t^{nz-1}\left(e^{ibt^n}-e^{-ibt^n}\right)dt\right]$$
Rewriting gives: $$\frac{1}{2i}\int_0^\infty e^{(ib-a)t^n}t^{nz-1}dt-\frac{1}{2i}\int_0^\infty e^{(-ib-a)t^n}t^{nz-1}dt$$
Then we do this substitution: \begin{align} c&=\pm ib-a\\ u&=-ct^n\\ du&=-nct^{n-1}dt \end{align} and rewrite the integral \begin{align} \int_0^\infty e^{ct^n}t^{nz-1}dt&=\int_0^\infty e^{ct^n}(t^n)^{z-1}t^{n-1}dt\\ &=\int_0^\infty e^{-u}\left(-\frac{u}{c}\right)^{z-1}\frac{-1}{nc}du\\ &=\frac{-1}{nc}\left(\frac{-1}{c}\right)^{z-1}\int_0^\infty e^{-u} u^{z-1}du\\ &=\frac{\Gamma(z)}{n}(-c)^{-z}\\ \end{align}
Adding everything together and rewriting $a-ib=\sqrt{a^2+b^2}e^{-i\theta}$ and $a+ib=\sqrt{a^2+b^2}e^{i\theta}$:
\begin{align} I&=\frac{\Gamma(z)}{2in}\left[(a-ib)^{-z}-(a+ib)^{-z}\right]\\ &=\frac{\Gamma(z)}{2in}\left[(\sqrt{a^2+b^2}e^{-i\theta})^{-z}-(\sqrt{a^2+b^2}e^{i\theta})^{-z}\right]\\ &=\frac{\Gamma(z)}{2in(a^2+b^2)^{z/2}}\left[(e^{-i\theta})^{-z}-(e^{i\theta})^{-z}\right]\\ &=\frac{\Gamma(z)}{2in(a^2+b^2)^{z/2}}\left[e^{iz\theta}-e^{-iz\theta}\right]\\ &=\frac{\Gamma(z)\sin(z\theta)}{n(a^2+b^2)^{z/2}} \end{align}