Integrals

**chiph588@** · September 23rd 2010, 10:48 AM

Show $\displaystyle \int_0^\infty \cos(x^k)dx = \Gamma\left(\frac{k+1}k\right)\cos\left(\frac\pi{2 k}\right)$ and $\displaystyle \int_0^\infty \sin(x^k)dx = \Gamma\left(\frac{k+1}k\right)\sin\left(\frac\pi{2 k}\right)$ .

**chisigma** · September 23rd 2010, 10:21 PM

Let's start from the general formula You can find in any 'Manual of Math' valid for $n>1$ ...

$\displaystyle \int_{0}^{\infty} \frac{dx}{x^{n} + a^{n}} = \pi\ \frac{a^{1-n}}{n\ \sin \frac{\pi}{n}}$ (1)

Now we define...

$\displaystyle \chi (t) = \int_{0}^{\infty} \cos (x^{k}\ t)\ dx$ (2)

... and compute...

$\displaystyle \mathcal{L} \{\chi(t)\} = \int_{0}^{\infty} e^{-s\ t}\ \int_{0}^{\infty} \cos (x^{k}\ t)\ dx\ dt = \int_{0}^{\infty} dx\ \int_{0}^{\infty} e^{-s\ t} \cos (x^{k}\ t)\ dt =$

$\displaystyle = \int_{0}^{\infty} \mathcal{L} \{\cos (x^{k}\ t)\}\ dx =\int_{0}^{\infty} \frac{s}{s^{2} + x^{2k}}\ dx$ (3)

From (3) and (1), setting $n=2\ k$ and $a=s^{\frac{1}{k}}$ we obtain...

$\displaystyle \mathcal{L} \{\chi(t)\} = \pi\ \frac{s^{\frac{1}{k} -1}}{2\ k\ \sin \frac{\pi}{2k}}$ (4)

... from which we derive $\chi(t)$ ...

$\displaystyle \chi(t) = \frac{\pi\ t^{-\frac{1}{k}}}{2\ k\ \Gamma(1-\frac{1}{k})\ \sin \frac{\pi}{2 k}}$ (5)

... and finally, setting $t=1$ in (5), we obtain...

$\displaystyle \int_{0}^{\infty} \cos x^{k}\ dx = \frac{\pi}{2\ k\ \Gamma(1-\frac{1}{k})\ \sin \frac{\pi}{2 k}}$ (6)

At first it seems that the result (6) is not correct

... but using the fact that...

$\displaystyle \frac{1}{\Gamma(1-\frac{1}{k})} = \frac{\sin \frac{\pi}{k}\ \Gamma(\frac{1}{k})}{\pi}$

$\displaystyle \Gamma(\frac{1}{k}) = k\ {\Gamma (1+\frac{1}{k})$

$\displaystyle \sin \frac{\pi}{k} = 2\ \sin \frac{\pi}{2k} \cos \frac{\pi}{2k}$

... You find that (6) is equivalent to the expression given by chiph588