While reading a paper, I experienced that my calculation results kept differing from the author's calculation results.

The authors of the paper seem to believe that the following equation holds:
$$\int_\mathbb{R} xf'(x) \Lambda^{2s}f dx= -\int_\mathbb{R} (\Lambda^s f)^2 dx-\frac{1}{2}\int_\mathbb{R}x[(\Lambda^s f)^2]' dx$$
where $s>3$ is a constant and $\Lambda^s f:=[(1+\xi^2)^{\frac{1}{2}}\hat{f}]^{\vee}$  , $\hat{f}(\xi):=\int_\mathbb{R}f(x) e^{-2\pi i\xi x}dx$ (Fourier transform) and $\check{f}(x):=\int_\mathbb{R} f(\xi)e^{2\pi i \xi x}dx$ (Inverse Fourier transform). $f$ is in Schwartz space $S(\mathbb{R})$.

However, my result is different with above. Following all lemmas and theorem are just my opinions.

Lemma 1. $\Lambda^s f=[(1+\xi^2)^{\frac{1}{2}}\hat{f}]^{\vee}=[(1+\xi^2)^{\frac{1}{2}}\check{f}]^{\wedge}$

proof) This hold because of change of variable $\xi \rightarrow -\xi$.
$$\Lambda^s f(x)=[(1+\xi^2)^{\frac{1}{2}}\hat{f}]^{\vee}=\int_\mathbb{R}(1+\xi^2)^{\frac{s}{2}}e^{2\pi i\xi x}\Big{(}\int_\mathbb{R}f(y)e^{-2\pi i\xi y}dy\Big{)}d\xi$$
$$=\int_\mathbb{R}(1+\xi^2)^{\frac{s}{2}}e^{-2\pi i\xi x}\Big{(}\int_\mathbb{R}f(y)e^{2\pi\xi y}dy\Big{)} d\xi=[(1+\xi^2)^{\frac{1}{2}}\check{f}]^{\wedge}\blacksquare$$

Lemma 2. $\int_\mathbb{R}\Lambda^sg\Lambda^sfdx=\int_\mathbb{R}g\Lambda^{2s}(f)dx$.

proof) This is results derived from weak Parseval's theorem i.e. 
$$\forall f,g\in S(\mathbb{R})\ \ \ \ \int_\mathbb{R}\hat{f}gdx=\int_\mathbb{R}f\hat{g}dx$$
If we use weak Parseval's theorem,
$$\int_\mathbb{R}\Lambda^s{f}\Lambda^s{g}dx=\int_\mathbb{R}[(1+\xi^2)^{\frac{s}{2}}\hat{f}]^{\vee}[(1+\xi^2)^{\frac{s}{2}}\check{g}]^{\wedge}dx=\int_\mathbb{R}\hat{f}(1+\xi^2)^s\check{g}d\xi=\int_\mathbb{R}[(1+\xi^2)^s\hat{f}]^{\vee}(\check{g})^{\wedge}dx$$
$$=\int_\mathbb{R}g\Lambda^{2s}f dx \blacksquare$$

Lemma 3. $\frac{d}{dx}(\Lambda^sf)(x)=\Lambda^s\Big{(}\frac{df(x)}{dx}\Big{)}$. In other words, Linear operator $\Lambda^s$ and $\frac{d}{dx}$ are commutative.

proof) In this proof, we use well-known facts about Fourier transform
$$2\pi i\xi \hat{f}(\xi)=(\frac{d}{dx}f)^{\wedge}(\xi),\ \ \ \frac{d}{d\xi}\hat{f}(\xi)=(-2\pi i) (xf(x))^\wedge(\xi).$$

If we use above facts,
$$\Lambda^s\Big{(}\frac{df(x)}{dx}\Big{)}=[(1+\xi^2)^{\frac{s}{2}}(f')^\wedge]^{\vee}=2\pi i [(1+\xi^2)^{\frac{s}{2}}\xi\hat{f}]^{\vee}$$
Then, 
$$\frac{d}{dx}(\Lambda^sf)(x)=\frac{d}{dx}\int_{\mathbb{R}}(1+\xi^2)^{\frac{s}{2}}e^{2\pi i\xi x}\hat{f}(\xi)d\xi=2\pi i [(1+\xi^2)^{\frac{s}{2}}\xi\hat{f}]^{\vee}=\Lambda^s\Big{(}\frac{df(x)}{dx}\Big{)}\blacksquare$$

$\bf Theorem$ $\int_\mathbb{R}xf'\Lambda^{2s}fdx=-\frac{1}{2}\int_\mathbb{R}x[(\Lambda^s f)^2]'dx+s\int_\mathbb{R}(\Lambda^sf)^2dx
-s\int_{\mathbb{R}}(\Lambda^{s-1}f)^2dx$ holds

If we use integration by parts and Lemma 3
$$\int_\mathbb{R}xf'(x)\Lambda^{2s} f dx=\Big{[}xf(x)\Lambda^{2s} f\Big{]}_{-\infty}^\infty-\int_\mathbb{R}xf(x)(\Lambda^{2s}f)'dx=-\int_\mathbb{R}xf(x)\Lambda^{2s}(f')dx$$

If we use Lemma 2 above is same with
$$=-\int_\mathbb{R}\Lambda^s(xf)\Lambda^s(f')dx$$
Meanwhile, if we use $(h')^{\vee}=-2\pi ix\check{h}$
$$x\Lambda^sf=x[(1+\xi^2)^{\frac{s}{2}}\hat{f}(\xi)]^{\vee}=\frac{i}{2\pi}[\frac{d}{d\xi}((1+\xi^2)^{\frac{s}{2}}\hat{f})]^{\vee}$$
$$=\frac{i}{2\pi}[s\xi(1+\xi^2)^{\frac{s-2}{2}}\hat{f}+(1+\xi^2)^{\frac{s}{2}}\hat{f}'(\xi)]^{\vee}$$
$$=\frac{i}{2\pi}[s\xi(1+\xi^2)^{\frac{s-2}{2}}\hat{f}]^{\vee}+[(1+\xi^2)^{\frac{s}{2}}(xf)^{\wedge}]^{\vee}$$
$$=\frac{i}{2\pi}[s\xi(1+\xi^2)^{\frac{s-2}{2}}\hat{f}]^{\vee}+\Lambda^s(xf)$$
In other words,
$$\Lambda^s(xf)=x\Lambda^sf-\frac{i}{2\pi}[s\xi(1+\xi^2)^{\frac{s-2}{2}}\hat{f}]^{\vee}$$
Therefore,
$$-\int_\mathbb{R}\Lambda^s(xf)\Lambda^s(f')dx=-\int_{\mathbb{R}}x\Lambda^sf\Lambda^s(f')dx
+\int_\mathbb{R}\frac{i}{2\pi}[s\xi(1+\xi^2)^{\frac{s-2}{2}}\hat{f}]^{\vee}\Lambda^s(f')dx$$
If we focus on the second term,
$$\frac{i}{2\pi}\int_\mathbb{R}[s\xi(1+\xi^2)^{\frac{s-2}{2}}\hat{f}]^{\vee}\Lambda^s(f')dx
=\frac{i}{2\pi}\int_\mathbb{R}[s\xi(1+\xi^2)^{\frac{s-2}{2}}\hat{f}]^{\vee}[(1+\xi^2)^{\frac{s}{2}}(f')^{\vee}]^{\wedge}dx$$
If we use weak Parseval theorem and $(h')^{\vee}=-2\pi ix\check{h}$,
$$=\int_\mathbb{R}s\xi^2(1+\xi^2)^{s-1}\hat{f}\check{f}d\xi
=\int_\mathbb{R}s(1+\xi^2)^s\hat{f}\check{f}d\xi-\int_\mathbb{R}s(1+\xi^2)^{s-1}\hat{f}\check{f}d\xi$$
$$=\int_{\mathbb{R}}s(\Lambda^sf)^2dx-\int_{\mathbb{R}}s(\Lambda^{s-1}f)^2dx$$

Therefore,
$$\int_\mathbb{R}xf'(x)\Lambda^{2s}fdx=-\int_\mathbb{R}x\Lambda^sf\Lambda^s{f'}dx+
\frac{i}{2\pi}\int_\mathbb{R}[s\xi(1+\xi^2)^{\frac{s-2}{2}}\hat{f}]^{\vee}\Lambda^s(f')dx$$
$$=-\frac{1}{2}\int_\mathbb{R}x[(\Lambda^s f)^2]'dx+s\int_\mathbb{R}(\Lambda^sf)^2dx
-s\int_{\mathbb{R}}(\Lambda^{s-1}f)^2dx \blacksquare$$

I wonder if the calculations of the paper authors are correct or if my calculations are correct. If anyone has any ideas for the calculations, any help would be greatly appreciated.