Finding the value of improper integral given some other integral's value Calculate the value of I. $$I=\int_0^{\infty} \frac{\sin(x)}{x^p}dx$$. Where 
$0<p<1,\int_0^{\infty}\frac{x^{p-1}}{1+x}dx=\frac{\pi}{\sin(p\pi)}$. 
$$\text{Attempt}$$ No concrete progress at all to be honest. Just tried doing $I(s)=\int e^{-sx^p}\frac{\sin(x)}{x^p}dx$ and then calculating $\frac{dI(s)}{ds}$ . But didn't help much . Another idea was calculating definite integral of $\frac{e^{ix}}{x^p}$ and then extracting the value of imaginary part of this integral.
 A: Using Laplace transform, we get
$$I=\int_0^\infty\mathscr{L}_t(\sin x)\mathscr{L}^{-1}_t(x^{-p})dt\\
=\int_0^\infty\frac{t^{p-1}}{1+t^2}\frac1{\Gamma(p)}dt$$
We can easily find $$\int_0^\infty\frac{t^{p-1}}{1+t^2}dt$$ by using substitution $u=t^2$:$$\int_0^\infty\frac{t^{p-1}}{1+t^2}dt=\frac12\int_0^\infty\frac{u^{p/2-1}}{1+u}du=\frac{\pi}2\csc\frac{p\pi}2$$
Therefore, $$I=\frac{\pi}2\csc\frac{p\pi}2\frac1{\Gamma(p)}= \Gamma (1-p)\cos\frac{p\pi}{2}$$
When $0<p<2$.
A: 
As mentioned within the comments this integral can be tackled utilizing Ramanuajan's Master Theorem as it was similarly done in this answer.

To actually apply Ramanujan's Master Theorem we have to reshape the integral a little bit. To be precise enforcing the substitution $x^2=u$ yields to the following
$$\begin{align}
\mathfrak{J}=\int_0^{\infty}\sin(x)x^{-p}~dx&=\int_0^{\infty}x^{-p}\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}dx\\
&=\frac12\int_0^{\infty}x^{-p}\sum_{n=0}^{\infty}\frac{(-x^2)^n}{(2n+1)!}2xdx\\
&=\frac12\int_0^{\infty}u^{-p/2}\sum_{n=0}^{\infty}\frac{(-u)^n}{(2n+1)!}du\\
&=\frac12\int_0^{\infty}u^{-p/2}\sum_{n=0}^{\infty}\frac{\Gamma(n+1)/\Gamma(2n+2)}{n!}(-u)^ndu
\end{align}$$
The last integral can be evaluated by applying Ramanujan's Master Theorem with $s=1-\frac p2$ and $\phi(n)=\frac{\Gamma(n+1)}{\Gamma(2n+2)}$. From hereon we further get
$$\begin{align}
\mathfrak{J}=\frac12\int_0^{\infty}u^{-p/2}\sum_{n=0}^{\infty}\frac{\Gamma(n+1)/\Gamma(2n+2)}{n!}(-u)^ndu&=\frac12\Gamma\left(1-\frac p2\right)\frac{\Gamma\left(-\left((1-\frac p2\right)+1\right)}{\Gamma\left(-2\left(1-\frac p2\right)+2\right)}\\
&=\frac1{2\Gamma(p)}\Gamma\left(\frac p2\right)\Gamma\left(1-\frac p2\right)\\
&=\frac1{2\Gamma(p)}\frac{\pi}{\sin\left(\frac{\pi p}{2}\right)}
\end{align}$$
Overall we can write down the equality

$$\mathfrak{J}=\int_0^{\infty}\sin(x)x^{-p}~dx=\frac1{2\Gamma(p)}\frac{\pi}{\sin\left(\frac{\pi p}{2}\right)}$$

The propesed representation of the integral invoking $\cos\left(\frac{\pi p}2\right)$ can be deduced farily easy by using Euler's Reflection Formula with $z=p$
$$\color{red}{\frac1{2\Gamma(p)}\frac{\pi}{\sin\left(\frac{\pi p}{2}\right)}}=\frac{\pi}{\Gamma(p)}\frac{\cos\left(\frac{\pi p}2\right)}{2\sin\left(\frac{\pi p}{2}\right)\cos\left(\frac{\pi p}2\right)}=\frac{\pi}{\Gamma(p)\sin(\pi p)}\cos\left(\frac{\pi p}2\right)=\color{red}{\Gamma(1-p)\cos\left(\frac{\pi p}2\right)}$$
A: by taking laplace transform of 
\begin{equation}
\int_{0}^{\infty}{\frac{\sin(x)}{x^p}}dx=\frac{1}{\Gamma{(p)}}\int_{0}^{\infty} \frac{s^{p-1}}{s^2+1}ds=\Gamma{(1-p)}\cos{\frac{p\pi}{2}}          
\end{equation}
A: NOT A SOLUTION:
As part of my work on another problem  I have been faced with the same problem. I have scoured and have only found a HyperGeometric representation for the integral. It is convergent for $p > 1$, $p \in \mathbb{R}$
