# Complex integration over segments and arc

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.

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

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

• Maybe you should explain why the integral is still from $0$ to $\infty$ after the substitution. – Marco Cantarini Jan 3 '17 at 10:55
• @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/… – DonAntonio Jan 3 '17 at 12:15