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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.

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  • $\begingroup$ One could apply Ramanujan's Master Theorem. The evaluation of the related integral $\int_0^{\infty} \frac{\cos(x)}{x^p}dx$ using this technique can be found here. $\endgroup$
    – mrtaurho
    Dec 23, 2018 at 12:01

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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 $0<p<2$.

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  • $\begingroup$ Thanks for the answer. $\endgroup$ Dec 23, 2018 at 10:16
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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 $z=p$

$$\color{red}{\frac1{2\Gamma(p)}\frac{\pi}{\sin\left(\frac{\pi p}{2}\right)}}=\frac{\pi}{\Gamma(p)}\frac{\cos\left(\frac{\pi p}2\right)}{2\sin\left(\frac{\pi p}{2}\right)\cos\left(\frac{\pi p}2\right)}=\frac{\pi}{\Gamma(p)\sin(\pi p)}\cos\left(\frac{\pi p}2\right)=\color{red}{\Gamma(1-p)\cos\left(\frac{\pi p}2\right)}$$

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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}

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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 $p > 1$, $p \in \mathbb{R}$

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    $\begingroup$ If it's not a solution, then it would be better to post it as a comment $\endgroup$
    – Jakobian
    Dec 23, 2018 at 10:10
  • $\begingroup$ Well it is a solution per sae, just not something that requires deviation. If fits the exact HyperGeometric function I linked. Do you think it requires some of the basic algebra required so the skilled mathematicians on the page will be able to follow>? $\endgroup$
    – user150203
    Dec 23, 2018 at 10:11
  • $\begingroup$ I do apologise for any semantic, spelling or basic grammar errors - my glasses are cracked :( $\endgroup$
    – user150203
    Dec 23, 2018 at 10:12

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