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I have seen that both the integrals $$ \color{black}{ \int_0^\infty \operatorname{sinc}(x^n)\,\mathrm{d}x = \frac{1}{n-1} \cos\left( \frac{\pi}{2n}\right)\Gamma \left(\frac{1}{n} \right) } $$ and $$ \color{black}{ \int_0^\infty \operatorname{sinc}^n(x)\,\mathrm{d}x = \frac{\pi}{2^n (n-1)!} \sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k {n \choose k} (n-2k)^{n-1} } $$ beeing computed here. Is there a similar expression for the more general $$ \int_0^\infty \frac{\sin^m(x^n)}{x^p}\,\mathrm{d}x $$

sine integral? I do not know precielly the restriction on the constant. I assume that they have to be positive. From my research $m$ and $p$ does not neccecary have to be equal, for example $$ \int_0^\infty \frac{\sin^3(x)}{x^2}\,\mathrm{d}x = \frac{4}{3} \log 3 \qquad \text{and} \qquad \int_0^\infty \frac{\sin^3(x^2)}{x^2}\,\mathrm{d}x = \frac{1}{4} \sqrt{\frac{\pi}{2}}\left( \sqrt{3} - 3\right) $$ but any closer restrictions on $n,m$ and $p$ I have not been able to gather.

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Through the substitution $x=z^{1/m}$ the problem boils down to finding $$ I(m,\alpha)=\int_{0}^{+\infty}\frac{\sin(x)^m}{x^\alpha}\,dx $$ and assuming $m\in\mathbb{N}$ we have that $\sin(x)^m$ can be expressed as a finite Fourier sine/cosine series.
So the problem boils down to finding, through the Laplace (inverse) transform, $$ \int_{0}^{+\infty}\frac{\sin(x)}{x^\alpha}\,dx = \Gamma(1-\alpha)\cos\left(\frac{\pi\alpha}{2}\right),\qquad \int_{0}^{+\infty}\frac{\cos(x)}{x^\alpha}\,dx=\Gamma(1-\alpha)\sin\left(\frac{\pi\alpha}{2}\right)$$ then extend the range of validity of such identities through integration by parts.

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  • $\begingroup$ Hello Jack Please can possibly give some reference where I can find the last integrals? I don't if it is possible it in the books by I. S. Gradshteyn ? if yes in which part? $\endgroup$ – Guy Fsone Nov 7 '17 at 9:57
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    $\begingroup$ @GuyFsone: they are standard instances of the Mellin transform. $\endgroup$ – Jack D'Aurizio Nov 7 '17 at 13:46
  • $\begingroup$ Can you provide me with some references (Good Books) $\endgroup$ – Guy Fsone Nov 7 '17 at 15:19
  • $\begingroup$ @GuyFsone: I suggest you my notes and A Course of Modern Analysis by Whittaker&Watson. $\endgroup$ – Jack D'Aurizio Nov 7 '17 at 17:05
  • $\begingroup$ Thanks a lot. I will definitely read it $\endgroup$ – Guy Fsone Nov 7 '17 at 17:26
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One alternative approach (compared to the answer of Jack D'Aurizio) is to write $$\tag{1} I(m,\alpha) = \int_0^1 \zeta(\alpha,x) \, \sin(x)^m \, \mathrm{d} x,$$ where $\zeta(\alpha,x)$ denotes the Hurwitz zeta function. We can use the binomial theorem in the form $$\sin(x)^m = (2 i)^{-m} \sum_{k=0}^m (-1)^k \binom{m}{k} (-1)^k e^{ix(n-2k)}$$ in order to reduce the problem to determining the fourier representation of the Hurwitz zeta function.

The right-hand side of (1) can be extended analytically for any $\alpha \neq 1$. For $\mathrm{Re}(\alpha) <1$ one has (by using the well-known 'Hurwitz formula') the identity $$ \zeta(\alpha,x)= 2 (2\pi )^{\alpha-1} \Gamma (1-\alpha) \left( \sin \big(\tfrac {\pi \alpha}{2}\big) \sum _{k=1}^{\infty} \frac{\cos(2 \pi x k)}{k^{1-\alpha}}+\cos \big(\tfrac{\pi \alpha}{2}\big) \sum_{k=1}^{\infty} \frac{\sin(2\pi x k)}{k^{1-\alpha}} \right).$$ By analytic continuation this gives explicite formulas for (1) also for complex-values.

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