Let $P_k(x)$ to be the first $k+1$ terms of the Taylor expansion of $\cos(x)$, that is $$P_k(x) = \sum_{l = 0}^k (-1)^l x^{2l}/(2l)!.$$ For $\alpha>0$ and $\alpha\notin \mathbb{Z}$, I want to evaluate the sign of the following integration: $$\int_{0}^\infty \frac{\cos(x) - P_{[\alpha]}(x)}{x^{2\alpha+1}}dx,$$ where $[\alpha]$ is the largest integer not larger than $\alpha$.
I have done some simulations and it seems the sign is $(-1)^{[\alpha]+1}$, which is the sign of the dominant term of the remaining Taylor series, but not fully sure about that. I wish someone could come up a solution for this. Thanks a lot!
Too long for comment
You already know that $$\cos x=\sum_{n\geq0}\frac{(-1)^nx^{2n}}{(2n)!}$$
So naturally, $$\begin{align} R_m(x)&=\cos x-P_{m}(x)\\ &=\sum_{n\geq0}\frac{(-1)^nx^{2n}}{(2n)!}-\sum_{n=0}^{m}\frac{(-1)^nx^{2n}}{(2n)!}\\ &=\sum_{n\geq m+1}\frac{(-1)^nx^{2n}}{(2n)!} \end{align}$$ And when we define $$I(\alpha)=\int_0^\infty \frac{\cos x-P_{\lfloor \alpha \rfloor}(x)}{x^{2\alpha+1}}\mathrm dx$$ We see that $$I(\alpha)=\int_0^\infty R_{\lfloor\alpha\rfloor}(x)\frac{\mathrm dx}{x^{2\alpha+1}}$$ $$I(\alpha)=\sum_{n\geq \lfloor\alpha\rfloor+1}\frac{(-1)^n}{(2n)!}\int_0^\infty x^{2n-2\alpha-1}\mathrm dx$$ But I do not know if this method is valid, as the final integral does not seem to converge...