Evaluation of a Fresnel type integral.

Question

I was evaluating the following 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?

Answer 1

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!

Answer 2

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 integral in \eqref{main} is of course in the sense of improper Riemann integrals, that is $$\int^\infty_0e^{-ix^p}\,dx:=\lim_{b\rightarrow\infty}\int^b_0e^{-ix^p}\,dx$$ 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 $$ F(\theta)=\int^\infty_0 f(x,\theta)\,dx,\qquad |p\theta|<\frac{\pi}{2}$$

1. Claim: $$\begin{align}F(\theta)=\int^\infty_0 e^{-e^{ip\theta}x^p}\,dx=\Gamma(1+\tfrac1p) e^{-i\theta},\quad |\theta p|<\frac{\pi}{2}\tag{2}\label{two}\end{align}$$ 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 $|\theta p|<\lambda<\tfrac{\pi}{2}$, there exists a neighborhood $I$ of $\theta$ such that $$|\theta' p|<\lambda$$ for all $\theta'\in I$. By the mean value theorem,
$$ \frac{|f(x,\theta')-f(x,\theta)|}{|\theta'-\theta|}\leq pe^{-\cos(\lambda)x^p}x^p\in L_1([0,\infty)) $$ From \eqref{three} along with dominated convergence we obtain that $$ F'(\theta)=\int^\infty_0 \partial_\theta f(x,\theta)\,dx=-i\int^\infty_0 x\partial_xf(x,\theta)\,dx$$ 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.

2. Claim: $$\begin{align} F(\theta)=\int^\infty_0\frac{1-p}{pe^{ip\theta}x^p}(f(x,\theta)-1)\,dx,\quad|\theta p|<\frac{\pi}{2}\tag{4}\label{four} \end{align}$$ Applying integration by parts to the integral in \eqref{two} leads to $$\begin{align} \int^\infty_0\frac{pe^{ip\theta}x^{p-1}}{pe^{ip\theta}x^{p-1}}f(x,\theta)\,dx&=-\frac{1}{pe^{ip\theta}}\int^\infty_0 \frac{1}{x^{p-1}}\partial_xf(x,\theta)\,dx=-\frac{1}{pe^{ip\theta}}\int^\infty_0 \frac{1}{x^{p-1}} G'_\theta(x)\,dx\\ &=-\frac{1}{pe^{ip\theta}}\left(\lim_{\substack{a\rightarrow0\\ b\rightarrow\infty}} \frac{G_\theta(x)}{x^{p-1}}\Big|^b_a-\int^b_a \frac{1-p}{x^p}(f(x,\theta)-1)\,dx\right)\\ &=\int^\infty_0\frac{1-p}{pe^{ip\theta}x^p}(f(x,\theta)-1)\,dx \end{align}$$ 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 $|x|>1$ $$ \frac{G_\theta(x)}{x^{p-1}}\leq \frac{2}{x^{p-1}}\xrightarrow{x\rightarrow\infty}0$$

3. 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}.

4. 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 $$\begin{align} \widehat{\phi_\alpha}(s)=\int^\infty_0 e^{-2\pi ixs}x^{\alpha-1}\,dx =(2\pi |s|)^{-\alpha}\Gamma(\alpha)e^{-\tfrac{i\operatorname{sign}(s)\alpha\pi}{2}}\tag{6}\label{six} \end{align}$$