Integrating the Difference between Sine and its Maclaurin Series

Question

The integral in question:

Let $n$ be a nonnegative integer.

$$ \int_0^\infty \frac{1}{x^{2n+3}} \left ( \sin x - \sum_{k=0}^n \frac{(-1)^k x^{2k+1}}{(2k + 1)!} \right ) \mathrm{d}x $$

I was given this problem to try for fun from a professor about two months ago and haven't made a dent in it. If possible, I'd like someone to show how they get to their answer, but please keep in mind that I haven't learned contour integration or complex analysis, but know of a few tricks like Feynman's.

What I've tried:

Writing out the terms to integrate by segments, using the Cosine Maclaurin series and substituting (pi/2 -x) to get the Sine series in a different way, integration by parts (2n+3) times or such to simplify, and a lot more I'm probably forgetting.

I know this site is full of integral-calculating gods, so if this piques your interest please have an attempt and share below.

Answer 1

This can be done via Integration by Parts. $$ \begin{align} &\int_0^\infty\frac1{x^{2n+3}}\left(\sin(x)-\sum_{k=0}^n\frac{(-1)^kx^{2k+1}}{(2k+1)!}\right)\mathrm{d}x\\ &=\frac{-1}{(2n+1)(2n+2)}\int_0^\infty\frac1{x^{2n+1}}\left(\sin(x)-\sum_{k=0}^{n-1}\frac{(-1)^kx^{2k+1}}{(2k+1)!}\right)\mathrm{d}x\tag1\\ &=\frac{(-1)^{n+1}}{(2n+2)!}\int_0^\infty\frac1x\,\sin(x)\,\mathrm{d}x\tag2\\[3pt] &=\frac{(-1)^{n+1}}{(2n+2)!}\frac\pi2\tag3 \end{align} $$ Explanation:
$(1)$: integrate by parts twice
$(2)$: repeat $(1)$ $n$ more times
$(3)$: see $(9)$ from this answer

Answer 2

Hoping that you enjoy hypergeometric functions $$I_n=\int_0^\infty \frac{1}{x^{2n+3}} \left ( \sin x - \sum_{k=0}^n \frac{(-1)^k x^{2k+1}}{(2k + 1)!} \right ) \,dx$$

$$\sum_{k=0}^n \frac{(-1)^k x^{2k+1}}{(2k + 1)!}=\frac{(-1)^n x^{2 n+3} \, _1F_2\left(1;n+2,n+\frac{5}{2};-\frac{x^2}{4}\right)}{(2 n+3)!}+\sin (x)$$

$$I_n=\frac{(-1)^{n+1}}{\Gamma (2 n+4)}\int_0^\infty \, _1F_2\left(1;n+2,n+\frac{5}{2};-\frac{x^2}{4}\right)\,dx$$ $$I_n=(-1)^{n+1}\,\frac{ \pi}{2\,\Gamma (2 n+3)}$$