Evaluating a complex integral with the goniometric function I was evaluating a this complex integral via gamma function:
$\int_0^\infty \sin (x^p) \,dx$ $\;$for $p \gt 1$,
so I expressed it as an imaginary part of $\int_0^\infty \exp(-ix^p) \, dx$ $\;$for $p \gt 1$


*

*The formula of the gamma function is $\Gamma (z) = \int_0^\infty x^{z-1} e^{-x} \, dx  $


I used the substitution $-y^{1/p}=xi$, $\;$ $\;$ $dx= \frac 1 p y^{\frac{1}{p}-1}i \, dy$ $\;$ $\;$and $\;$ $\;$ $\frac {1}{p} = \alpha$
Then $\int_0^\infty \alpha i y^{\alpha-1}e^{-y} \, dx = \alpha i \Gamma (\alpha) = \ i \frac {1}{p} \Gamma (\frac {1}{p})$
The solution according to my textbook is $\  \frac {1}{p} \Gamma (\frac {1}{p}) \sin (\frac {\pi}{2p})$
But I think $\sin (\frac {\pi}{2p})$ is right if I have ${i}^p$, but I got just $i$.
My solution is then $\  \frac {1}{p} \Gamma (\frac {1}{p}) \sin (\frac {\pi}{2}) =\frac {1}{p} \Gamma (\frac {1}{p})$.
Did I miss something important?
EDIT
I tried to calculate this integral for $p = 2$ and the textbook is right, but why?

You can first substitute $u=x^p$:
\begin{align}
I=\int_0^\infty \sin (x^p)\,dx&=\frac{1}{p}\int_0^\infty u^{\frac{1}{p}-1}\sin u \,du \\ \\
&=\frac{1}{p} \Im\int_0^\infty u^{\frac{1}{p}-1} e^{iu}\,du \\ \\
&=\frac{1}{p}\Gamma\left(\frac{1}{p}\right) \Im i^{1/p} \\ \\
&=\frac{1}{p}\Gamma\left(\frac{1}{p}\right) \Im e^{\frac{i\pi}{2p}} \\ \\
&=\Gamma\left(1+\frac{1}{p}\right)\sin\frac{\pi}{2p}
\end{align}
You can also evaluate the integral via a useful property of the Laplace Transform:
\begin{align}
\int_0^\infty f(x)\,g(x)\,dx=\int_0^\infty \mathcal{L}^{-1}\{f(x)\}(s)\mathcal{L}\{g(x)\}(s)\,ds
\end{align}
Then, 
\begin{align}
\int_0^\infty \sin (x^p)\,dx&=\frac{1}{p}\int_0^\infty u^{\frac{1}{p}-1}\sin u \,du \\ \\
&=\frac{1}{p\,\Gamma\left(1-\frac{1}{p}\right)}\int_0^\infty \frac{s^{-\frac{1}{p}}}{s^2+1}\,ds \qquad s^2\mapsto u \\ \\
&=\frac{1}{2p\,\Gamma\left(1-\frac{1}{p}\right)} \int_0^\infty \frac{u^{-\frac{1}{2}(\frac{1}{p}+1)}}{1+u}\,du \\ \\
&=\frac{1}{2p\,\Gamma\left(1-\frac{1}{p}\right)} \mathcal{B}\left[\frac{1}{2}\left(1-\frac{1}{p}\right),\frac{1}{2}\left(1+\frac{1}{p}\right) \right] \\ \\
&= \frac{1}{2p\,\Gamma\left(1-\frac{1}{p}\right)} \Gamma\left(\frac{1}{2}\left(1-\frac{1}{p}\right)\right) \Gamma \left(\frac{1}{2}\left(1+\frac{1}{p}\right) \right) \\ \\
&=\frac{\pi}{2p\,\Gamma\left(1-\frac{1}{p}\right)}\sec\left(\frac{\pi}{2p} \right) \\ \\
&= \frac{\pi \sin \left(\frac{\pi}{2p} \right)}{p\,\sin \left(\frac{\pi}{p} \right)\,\Gamma\left(1-\frac{1}{p}\right)} \\ \\
&= \frac{1}{p} \Gamma \left(\frac{1}{p}\right) \sin \frac{\pi}{2p} \\ \\
&= \Gamma \left(\frac{1}{p}+1\right) \sin \frac{\pi}{2p}
\end{align}
Feel free to ask if you have any questions!
