Does this integral have a closed form or asymptotic expansion? $\int_0^\infty \frac{\sin(\beta u)}{1+u^\alpha} du$ I'm interested in this following definite integral:
$$
\int_0^\infty du \frac{\sin(\beta u)}{1+u^\alpha},
$$
where $\beta>0$ and $\alpha\geq1$.  Is there any closed form for this integral?  I would be fine if the answer involves special functions, it would just be nice to have the answer in closed form.  
I'm also interested in the behavior of this integral in the limit $\beta\gg1$.  In particular, I'd like to know if it decays exponentially at large values of $\beta$.  
 A: Assuming $\beta >0$, considering $$I_a=\int_0^\infty  \frac{\sin(b u)}{1+u^a}\,du$$ it seems that $a=1$ corresponds to a very particular case $$I_1=\text{Ci}(b) \sin (b)-\text{Si}(b) \cos (b)+\frac{1}{2} \pi  \cos (b)$$ where appear the sine and cosine integrals. $I_1$ seems to decrease as an hyperbola. Expanding $I_1$ for large values of $b$ up to $O\left(\frac{1}{b^9}\right)$ gives $$I_1\approx \frac{1}{b}-\frac{2}{b^3}+\frac{24}{b^5}-\frac{720}{b^7}+\cdots$$
Fo the other cases, using a CAS, it seems that $I_a$ systematically express in terms of the Meijer G function. For example 
$$I_2=\frac{1}{4} \sqrt{\pi } b G_{1,3}^{2,1}\left(\frac{b^2}{4}|
\begin{array}{c}
 0 \\
 0,0,-\frac{1}{2}
\end{array}
\right)$$
$$I_3=\frac{G_{1,7}^{5,1}\left(\frac{b^6}{46656}|
\begin{array}{c}
 \frac{5}{6} \\
 \frac{1}{6},\frac{1}{3},\frac{1}{2},\frac{5}{6},\frac{5}{6},0,\frac{2}{3}
\end{array}
\right)}{2 \sqrt{3 \pi }}$$ $$I_4=\frac{1}{2} \sqrt{\frac{\pi }{2}} G_{1,5}^{3,1}\left(\frac{b^4}{256}|
\begin{array}{c}
 \frac{3}{4} \\
 \frac{1}{4},\frac{3}{4},\frac{3}{4},0,\frac{1}{2}
\end{array}
\right)$$ I have not been able to see any clear pattern. 
For $b >1$, these functions start at $0$, go through a maximum value and seem to also decrease as hyperbolas at large values of $b$ (just as Taozi commented).
