1
$\begingroup$

For $0 < p < 1$, compute the integrals $$\int_0^\infty x^{p-1} \cos x dx, \;\;\; \int_0^\infty x^{p-1} \sin x dx$$

I tried by integrating the function $z^{(p-1)}e^{-z}$ over the union of the segment $[r,R]$, arc $e^{i \theta}$, $\theta \in [0,\pi/2]$, segment $[ir,iR]$ and arc $e^{i \theta}$, $\theta \in [\pi/2,0]$ but i am stuck with the integration.

Any other approach or any help will be truly appreciated.

$\endgroup$
1
$\begingroup$

An idea for you to complete:

Taking

$$\;\int_0^\infty x^{p-1}e^{-ix}dx=\int_0^\infty x^{p-1}\cos x\,dx+i\int_0^\infty x^{p-1}\sin x\,dx\;:$$

$$t=ix\implies \,-i\,dt=\frac{dt}i=dx\;,\;\;-i\int_0^\infty\left(-it\right)^{p-1}e^{-t}dt=(-i)^{p}\Gamma(p)$$

and now

$$(-i)^{p}=e^{p\,\text{Log(-i)}}=e^{p\left(\log1-i\frac\pi2\right)}=e^{-p\pi i/2}\,$$

Thus, for example

$$p=\frac12\implies\int_0^\infty x^{-1/2}\cos x=\text{Re}\,\left(e^{-\pi i/4}\Gamma\left(\frac12\right)\right)=\sqrt\frac\pi2$$

or also

$$p=\frac23\implies\int_0^\infty x^{-1/3}\cos x=\text{Re}\,\left(e^{-\pi i/3}\Gamma\left(\frac23\right)\right)=\frac{\Gamma\left(\frac23\right)}2$$

$\endgroup$
  • $\begingroup$ Maybe you should explain why the integral is still from $0$ to $\infty$ after the substitution. $\endgroup$ – Marco Cantarini Jan 3 '17 at 10:55
  • $\begingroup$ @MarcoCantarini Perhaps that's something the OP could complete, yet: when we integrate in some rectangle in the complex plane, the line integrals over the vertical sides vanish (hopefully), getting the desired equality. One can check the following, too: math.stackexchange.com/questions/648043/… $\endgroup$ – DonAntonio Jan 3 '17 at 12:15

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.