what is the answer of $$\int \limits_{0}^{\infty}\frac {\sin (x^n)} {x^n}dx$$

From this A sine integral $\int_0^{\infty} \left(\frac{\sin x }{x }\right)^n\,\mathrm{d}x$ I saw the answer for $$\int \limits_{0}^{\infty}\left(\frac {\sin x} {x}\right)^ndx$$

but for my question i didn't see any answer

is there any help

thanks for all


3 Answers 3



$$f(t)=\int_0^{\infty} \frac{\sin tx^n}{x^n}\,dx\Rightarrow f'(t)=\int_0^{\infty}\cos tx^n\,dx=\frac{1}{\sqrt[n]{t}}\int_0^{\infty}\cos x^n\,dx$$

This is the generalised Fresnel integral, which evaluates to:

$$ \int_0^{\infty} \cos x^n\,dx=\cos \left(\frac{\pi}{2n}\right) \Gamma \left(\frac{n+1}{n}\right)\;\;(\star)$$

Noting that $f(0)=0:$

$$f(1)=\cos \left(\frac{\pi}{2n}\right) \Gamma \left(\frac{n+1}{n}\right)\int_0^1\frac{1}{\sqrt[n]{t}}dt=\cos\left(\frac{\pi}{2n}\right)\Gamma \left(\frac{1}{n}\right)\frac{1}{n-1}$$

As requested, here is a proof $(\star):$ consider the following paths:

$$\begin{aligned}\gamma(x)=x,\, 0 \leq x\leq r,\;\;\gamma'(t)=te^{i \frac{\pi}{2n}},\, 0\leq t\leq r,\;\;\mu(\theta)=re^{i \theta},\, 0\leq \theta \leq \tfrac{\pi}{2n}\end{aligned}$$

By Cauchy, $\displaystyle \int_{\gamma} e^{iz^n}\,dz+\int_{\mu} e^{iz^n}\,dz=\int_{\gamma'} e^{iz^n}\,dz$

On the RHS, as $r\to\infty:$

$$\begin{aligned}\displaystyle\int_{\gamma'} e^{iz^n}\,dz=e^{i\frac{\pi}{2n}}\int_0^{r} e^{-t^n}dt=\frac{e^{i\frac{\pi}{2n}}}{n}\int_0^{\sqrt[n]{r}} s^{\frac{1}{n}-1}e^{-s}\,ds\to \frac{e^{i\frac{\pi}{2n}}}{n}\, \Gamma \left(\frac{1}{n}\right)=e^{i \frac{\pi}{2n}}\Gamma\left(\frac{n+1}{n}\right)\end{aligned}$$

On the LHS, as $r\to\infty:$

$$ \left|\int_{\mu} e^{iz^n}\,dz\right|= \left|ir\int_0^{\frac{\pi}{2n}}e^{i\left(r^n e^{in\theta}+\theta\right)}\,d \theta\right| \leq r\int_0^{\frac{\pi}{2n}} e^{-r^n\sin n\theta}\,d\theta \to 0$$

and obviously $\displaystyle \int_{\gamma} e^{iz^n}\,dz=\int_0^{\infty} e^{ix^n}\,dx$ as $r\to\infty$

Thus, equating real & imaginary parts:

$$ \int_0^{\infty} \cos x^n\,dx=\cos \left(\frac{\pi}{2n}\right) \Gamma \left(\frac{n+1}{n}\right)$$

$$ \int_0^{\infty} \sin x^n\,dx=\sin \left(\frac{\pi}{2n}\right) \Gamma \left(\frac{n+1}{n}\right)$$

  • $\begingroup$ thanks alot but how to prove that $$ \int_0^{\infty} \cos x^n\,dx=\cos \left(\frac{\pi}{2n}\right) \Gamma \left(\frac{n+1}{n}\right)$$ $\endgroup$
    – mnsh
    Jul 28, 2013 at 2:07
  • 1
    $\begingroup$ @hmedan.mnsh Added ;) $\endgroup$
    – L. F.
    Jul 28, 2013 at 2:09

Using the change of variables $x^n=t$ gives

$$ I = \int \limits_{0}^{\infty}\frac {\sin (x^n)} {x^n}dx=\frac{1}{n}\int _{0}^{\infty }\!\sin\left( t \right) {t}^{{\frac {1-2\,n}{n}}} {dt}.$$

Now, recalling the Mellin transform of a function $f$

$$ F(s)=\int_{0}^{\infty}x^{s-1}f(x) dx. $$

One can right away evaluate $I$ by finding the Mellin transform of $\sin(x)$ which is

$$\Gamma \left( s \right) \sin \left( \frac{\pi \,s}{2} \right), $$

and then substituting $s=\frac{1-n}{n}$ (since $s-1={\frac {1-2\,n}{n}}$ ) to get

$$I = \frac{1}{n} \Gamma\left( \frac{1-n}{n} \right) \sin\left(\frac{\pi}{2}\frac{1-n}{n} \right)\quad n\in \mathbb{N}. $$

Note: The following is a useful identity

$$ \Gamma(1-z) \Gamma(z) = {\pi \over \sin{(\pi z)}}, $$

which is known as the Euler's reflection formula.


Start by integration by parts

$$ I= \frac{n}{n-1}\int^{\infty}_0 \cos(x^n)\, dx$$

Now use the transformation $x^n \to x $

$$ I= \frac{1}{n-1}\int^{\infty}_0 \cos(x)\, x^{\frac{1}{n}-1}\, dx$$

Notice that

$$\Re(I) = \Re \left( \frac{1}{n-1}\int^{\infty}_0 e^{-ix}\, x^{\frac{1}{n}-1}\,dx\right)$$

Consider the Laplace transform

$$ \Re\left( \frac{1}{n-1} \frac{\Gamma \left(\frac{1}{n} \right)}{\sqrt[n]{-i}} \right)$$

Using the principle logarithm

$$\Re\left( \frac{e^{\frac{\pi}{2n}i}}{n-1} \Gamma \left(\frac{1}{n} \right) \right)$$

Hence we have

$$I = \frac{1}{n-1} \cos\left( \frac{\pi}{2n}\right)\Gamma \left(\frac{1}{n} \right) $$

For a more general solution look at http://integralsandseries.prophpbb.com/topic6.html

  • $\begingroup$ but how they prove the general case $\endgroup$
    – mnsh
    Jul 28, 2013 at 18:55
  • $\begingroup$ Hey , nice to see arabic people here. What general case ? $\endgroup$ Jul 28, 2013 at 19:51
  • $\begingroup$ nice to see you وانا ارسلت لك طلب عالفايسبوك I mean the general case in your link $\endgroup$
    – mnsh
    Jul 28, 2013 at 20:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.