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?
 A: 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!
A: Here is an alternative solution based on s series of problems in Jones, F., Lebesgue Integration in Euclidean Space, Jones & Bartlett Publishers, 2001, Problems 31, 32 and 33, pp 249-250.  We prove that
$$ \begin{align}
\int^\infty_0 e^{-ix^p}\,dx &= \Gamma(1+\tfrac1p)e^{-\tfrac{i\pi}{2p}}
\tag{0}\label{main}\end{align}
$$
where $p>1$. The solution to the particular problem in the OP follows by separating the real and imaginary parts of our main identity \eqref{main}, that is
$$\begin{equation}
\begin{split}
\int^\infty_0\cos(x^p)\,dx&=\Gamma(1+\tfrac1p)\cos(\tfrac{\pi}{2p})\\
\int^\infty_0\sin(x^p)\,dx&=\Gamma(1+\tfrac1p)\sin(\tfrac{\pi}{2p})
\end{split}\tag{1}\label{one}
\end{equation}
$$
The rest of this answer is dedicated to prove \eqref{main}.
We assume throughout that $p>1$. Define the function $f:[0,\infty)\times(-\tfrac{\pi}{2p},\tfrac{\pi}{2p})\rightarrow\mathbb{R}$
$$ f(x,\theta)=e^{-e^{i\theta p}x^p}$$
For any $\theta$ with $|\theta p|<\frac{\pi}{2}$, $f(\cdot,\theta)\in L_1((0,\infty),\mathscr{B}((0,\infty)),\lambda)$ since $\cos(\theta p)>0$, $p>1$, and
$$|f(x,\theta)|=e^{-\cos(\theta p) x^{p}}$$
Define
$$ F(\theta)=\int^\infty_0 f(x,\theta)\,dx,\qquad |p\theta|<\frac{\pi}{2}$$

*

*Claim:
$$\begin{align}F(\theta)=\int^\infty_0 e^{-e^{ip\theta}x^p}\,dx=\Gamma(1+\tfrac1p) e^{-i\theta}\tag{2}\label{two}\end{align}$$
A simple calculation shows that
$$\begin{align} \partial_\theta f =-ip x^p e^{i\theta p} f(x,\theta)=ix \partial_xf\tag{3}\label{three}\end{align}$$
Now, for $\theta\in(-\tfrac{\pi}{2p},\tfrac{\pi}{2p})$ fixed, and any $\tfrac{\pi}{2}-|\theta p|<\lambda<\tfrac{\pi}{2}$, there exists a neighborhood $I$ of $\theta$ such that $$|\theta' p|<\lambda$$ for all $\theta'\in I$. By the mean value theorem,
$$
\frac{|f(x,\theta')-f(x,\theta)|}{|\theta'-\theta|}\leq pe^{-\cos(\lambda)x^p}x^p\in L_1([0,\infty))
$$
Applying dominated convergence we obtain that
$$
F'(\theta)=\int^\infty_0 \partial_\theta f(x,\theta)\,dx=-i\int^\infty_0 x\partial_xf(x,\theta)\,dx$$
by  \eqref{three}. Setting $G_\theta(x):=\int^x_0 \partial_1f(u,\theta)\,du=f(x,\theta)-f(0,1)=f(x,\theta)-1$ and applying integration by parts we obtain
$$\begin{align}
F'(\theta)&=i\int^\infty_0 x G'_\theta(x)\,dx=i\lim_{\substack{a\rightarrow0\\ b\rightarrow\infty}}\left(xG_\theta(x)|^b_a-\int^b_aG_\theta(x)\,dx\right)\\
&= i\lim_{\substack{a\rightarrow0\\ b\rightarrow\infty}}\left(bf(b,\theta) -b-af(a,\theta)+a -\int^b_af(x,\theta)\,dx-(b-a)\right)\\
&=i\lim_{\substack{a\rightarrow0\\ b\rightarrow\infty}}\left( bf(b,\theta)-af(a,\theta)-\int^b_a f(x,\theta)\,dx\right)\\
&=-i\int^\infty_0 f(x,\theta)\,dx=-i F(\theta)
\end{align}
$$
Notice that the limit in the integral is valid by  dominated convergence. Hence $F(\theta)=F(0)e^{-i\theta}$. On the other hand
$$F(0)=\int^\infty_0 e^{-x^p}\,dx=\frac{1}{p}\int^\infty_0 e^{-u}u^{\tfrac1p-1}\,du=\Gamma(1+\tfrac1p).$$
This concludes the proof of the claim.


*Claim:
$$\begin{align}
F(\theta)=\int^\infty_0\frac{1-p}{pe^{ip\theta}x^p}(f(x,\theta)-1)\,dx\tag{4}\label{four}
\end{align}$$
Applying integration by parts to the integral in \eqref{two} leads to
$$\begin{align}
\int^\infty_0\frac{pe^{ip\theta}x^p}{pe^{ip\theta}x^p}f(x,\theta)\,dx&=-\frac{1}{pe^{ip\theta}}\int^\infty_0 \frac{1}{x^{p-1}}\partial_xf(x,\theta)\,dx=-\frac{1}{pe^{ip\theta}}\int^\infty_0 \frac{1}{x^{p-1}} G'_\theta(x)\,dx\\
&=-\frac{1}{pe^{ip\theta}}\left(\lim_{\substack{a\rightarrow0\\ b\rightarrow\infty}} \frac{G_\theta(x)}{x^{p-1}}\Big|^b_a-\int^b_a \frac{1-p}{x^p}(f(x,\theta)-1)\,dx\right)\\
&=\int^\infty_0\frac{1-p}{pe^{ip\theta}x^p}(f(x,\theta)-1)\,dx
\end{align}$$
The limits in the integration are taken in the sense of Lebesgue which is valid by dominated convergence. The last identity follows from the fact that:


*

*(a) $|e^z-1-z|\leq |z|^2$ for all $|z|\leq1$. This implies that $|e^z-1|\leq |z|+|z|^2$. When $0\leq x\leq 1$, setting $z=-e^{ip\theta}x^p$ gives
$$\Big|\frac{G_\theta(x)}{x^{p-1}}\Big|=\frac{|e^{-e^{ip\theta}x^p}-1|}{x^{p-1}}\leq |x|+|x|^{p+1}\xrightarrow{x\rightarrow0}0$$

*(b) For $|x|>1$
$$ \frac{G_\theta(x)}{x^{p-1}}\leq \frac{2}{x^{p-1}}\xrightarrow{x\rightarrow\infty}0$$


*Proof of main identity \eqref{main}:
First notice that
$$\Big|\frac{f(x,\theta)-1}{x^p}\Big|\leq (1+x^p)\mathbb{1}_{[0,1]}(x) +\frac{2}{x^p}\mathbb{1}_{(1,\infty)}(x)\in L_1((0,\infty)).$$
Then, an application of dominated convergence on \eqref{four} yields
$$\begin{align}
\lim_{\theta\rightarrow\frac{\pi}{2p} }F(\theta)=\int^\infty_0\frac{1-p}{ip x^p}(e^{-ix^p}-1)\,dx=\Gamma(1+\tfrac1p)e^{-\tfrac{i\pi}{2p}}\tag{5}\label{five}
\end{align}$$
Applying integration by parts to the integral in \eqref{five} gives
$$\begin{align}
\int^\infty_0\frac{1-p}{ip x^p}(e^{-ix^p}-1)\,dx&=\frac{1-p}{ip}\int^\infty_0 \frac{1}{x^p}(e^{-ix^p}-1)\,dx\\
&=\frac{1}{ip}\lim_{\substack{a\rightarrow0\\ b\rightarrow\infty}}\left( \frac{e^{-ix^p}-1}{x^{p-1}}\Big|^b_a +\int^b_a ip e^{-ix^p}\,dx \right)\\
&=\int^\infty_0 e^{-ix^p}\,dx
\end{align}$$
where the last integral converges in the sense of an improper Riemann integral. The identity \eqref{main} follows from \eqref{two} and \eqref{five}.


*An additional application:



*

*Taking the conjugate in \eqref{main} gives
$$\begin{align}
\int^\infty_0 e^{ix^p}\,dx=\Gamma(1+\tfrac1p)e^{\tfrac{i\pi}{2p}}\tag{0'}\label{mainp}\end{align}$$
Using the change pf variables $u=x^p$, and then $u=sv$, for $s>0$ in \eqref{main} and \eqref{mainp} we obtain
$$\int^\infty_0 e^{\mp ix^p}\,dx= \frac1p\int^\infty_0 e^{\mp iu}\frac{1}{u^{1-1/p}}\,du=\frac{s^{1/p}}{p}\int^\infty_0 e^{\mp svi}v^{\tfrac1p-1}\,dv$$
That is
$$ps^{-\tfrac1p}\Gamma(1+\tfrac1p)e^{\mp \tfrac{i\pi}{2p}}=\int^\infty_0 e^{\mp svi}v^{\tfrac1p-1}\,dv$$
Setting $0<\alpha=\tfrac1p<1$,  and  $\phi_\alpha(x):=x^{\alpha-1}\mathbb{1}_{(0,\infty)}(x)$, we have that the Fourier transform $\phi_\alpha$ is given by
$$\begin{align}
\widehat{\phi_\alpha}(s)=\int^\infty_0 e^{-2\pi ixs}x^{\alpha-1}\,dx =(2\pi |s|)^{-\alpha}\Gamma(\alpha)e^{-\tfrac{i\operatorname{sign}(s)\alpha\pi}{2}}\tag{6}\label{six}
\end{align}$$
