integral research solution (mellin transform) what is the result of this integral:
$$\int_{-\infty}^{+\infty}x^{-5/3} \cos \left[ \left(x-1 \right)\times h\right] \, \mathrm{d}x$$
with : h a constant.
Thank you.
 A: Just use the identity 
$$ \cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b) $$
then proceed just like in the previous problem.
Added: The approach which I gave you is correct. However you are now asking a different question. You are looking for a real value which Robert gave you an answer and we thank him for doing that. Let me answer it in a more general setting. 
Your integral is a special case of the more general integral 
$$I(\alpha) =\int_{-\infty}^{\infty}x^{\alpha} \cos((x-1)h) dx $$ $$
= \int_{0}^{\infty}x^{\alpha} \cos((x-1)h) dx  + (-1)^\alpha\int_{0}^{\infty}x^\alpha \cos((x+1)h) dx $$
 which is a multivalued function in general. In your case $\alpha = -\frac{5}{3}$ which causes the $(-1)^{-5/3}=(-1)^{1/3}$ (see complex variables) to have multi-values. If you are interested in a real value of the integral you can choose the right value which is in this case $-1$. So choosing the root $-1$ which gives you the answer
$$ I(-5/3) = \int_{0}^{\infty}x^{-5/3} \cos((x-1)h) dx  + (-1)\int_{0}^{\infty}x^{-5/3} \cos((x+1)h) dx $$
$$ I(-5/3) = {\frac {3\pi \,{h}^{2/3} }{\Gamma  \left( 2/3\right) }}\sin (h)$$
which agrees with Robert's answer.
Note: 
1) The roots of $z^3+1$ are

$$ -1,\, \frac{1}{2}-\frac{\sqrt{3}}{2},\, \frac{1}{2}+\frac{\sqrt{3}}{2}$$

2) You can do the same with the formula I gave you in the previous problem to get a real value of the integral.
A: If you want a real answer, you want to use the real cube root of $x$ when $x < 0$.
This is also needed to cancel the divergence of the integral at $0$ (using a Cauchy singular value).  So write the integral as
$$\eqalign{ &\int_0^\infty  x^{-5/3} \left(\cos(h (x-1)) - \cos(h (-x-1))\right)\; dx\cr 
&= 2 \sin(h) \int_0^\infty  x^{-5/3} \sin(hx)\; dx\cr
&= 2 h^{2/3} \sin(h) \int_0^\infty t^{-5/3} \sin(t)\; dt\cr
&= \dfrac{3\pi}{\Gamma(2/3)} h^{2/3} \sin(h) }$$
