What is $\int\limits_{0}^{\infty}\frac{\sin(x^n)}{x^2+1}dx$?

Question

Let $I_n=\int\limits_{0}^{\infty}\frac{\sin(x^n)}{x^2+1}dx$

Wolfram Alpha gives an exact value for $n=-4,-3,-2,-1,0,1,2,3,4,6,8$ and interestingly enough, it seems $I_n=I_{-n}$.

$$I_0=\frac{\pi}{2}\sin(1)$$ $$I_1=I_{-1}=\frac{\operatorname{Ei}(1)-e^2\operatorname{Ei}(-1)}{2e}$$ $$I_2=I_{-2}=\frac{\pi}{2}\left(\sin(1)\left(\operatorname{C}\left(\sqrt{\frac{2}{\pi}}\right)+\operatorname{S}\left(\sqrt{\frac{2}{\pi}}\right)\right)+\cos(1)\left(\operatorname{C}\left(\sqrt{\frac{2}{\pi}}\right)-\operatorname{S}\left(\sqrt{\frac{2}{\pi}}\right)\right)\right)$$ And so on... $$I_{\frac{1}{2}}=\frac{\pi}{e^\frac{1}{\sqrt{2}}}\sin\left(\frac{1}{\sqrt{2}}\right)$$ So my question is: For any $n$, what is $\int\limits_{0}^{\infty}\frac{\sin(x^n)}{x^2+1}dx$?

Answer 1

With help of Mathematica and using Mellin and Inverse Mellin transform:

$$\int_0^{\infty } \frac{\sin \left(x^n\right)}{x^2+1} \, dx=\\\mathcal{M}_s^{-1}\left[\mathcal{M}_x\left[\sin \left(x^n\right)\right](s) \mathcal{M}_x\left[\frac{1}{1+x^2}\right](1-s)\right](1)=\\\mathcal{M}_s^{-1}\left[\frac{\pi \sec \left(\frac{\pi s}{2}\right) \left(\Gamma \left(\frac{s}{n}\right) \sin \left(\frac{\pi s}{2 n}\right)\right)}{2 | n| }\right](1)=\\\mathcal{M}_s^{-1}\left[\frac{\pi \Gamma \left(\frac{1}{2}-\frac{s}{2}\right) \Gamma \left(\frac{s}{n}\right) \Gamma \left(\frac{1+s}{2}\right)}{2 | n| \Gamma \left(\frac{s}{2 n}\right) \Gamma \left(1-\frac{s}{2 n}\right)}\right](1)$$

Then using definition of FoxH function I get:

Mathematica code:

Pi/(2 Abs[n])*FoxH[{{{1,-(1/n)}}, {{0,1/(2 n)}}}, {{{1/2,-(1/2)}, {1/2, 1/2}}, {{0,1/(2 n)}}}, 1] for:$n>0$ and $n\in \mathbb{R}$,