Why this Equivalence of integrals is true? $$\int_b^{\infty}(1+y^{2})e^{-y^2}\left[\frac{f(y)-f(x)}{y-x}\right]^{2}dy=O(1)\int_b^{\infty}e^{-y^2}\left[f^2(y)+f^2(x)\right]dy$$
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go to pg 72 in the end  this pg. I don't understand how to get to the product of $O(1)$ by the integral. One idea for my problem.
What happen with $\int_{-b}^{b}(1+y)^2e^{-y^2}\varphi^2(x,y)dy$? converges? why?
thanks

The integrand can be rewritten as
$$
\begin{align}
&(1+y^{2})e^{-y^2}\left[\frac{f(y)-f(x)}{y-x}\right]^{2} \\
&\qquad = e^{-y^2}\left[f^2(y) - 2f(y)f(x) + f^2(x)\right] \frac{1+y^2}{(y-x)^2} \\
&\qquad = e^{-y^2}\left[f^2(y) + f^2(x)\right]\left[1 - \frac{2f(y)f(x)}{f^2(y)+f^2(x)}\right]\frac{1+y^2}{(y-x)^2}.
\end{align}
$$
Now
$$
\left|1 - \frac{2f(y)f(x)}{f^2(y)+f^2(x)}\right| \leq 1 + \left|\frac{2f(y)f(x)}{f^2(y)+f^2(x)}\right| \leq 2,
$$
which can be seen by maximizing the function
$$
x \mapsto \frac{2ax}{x^2+a^2}.
$$
Also
$$
\lim_{y\to\infty} \frac{1+y^2}{(y-x)^2} = 1
$$
if $x$ is fixed (as is the case in the paper), so the quantity $\frac{1+y^2}{(y-x)^2}$ is bounded for all $y \geq b > x$.
Thus
$$
\left[1 - \frac{2f(y)f(x)}{f^2(y)+f^2(x)}\right]\frac{1+y^2}{(y-x)^2} = O(1),
$$
and so
$$
\int_b^{\infty}(1+y^{2})e^{-y^2}\left[\frac{f(y)-f(x)}{y-x}\right]^{2}dy=O(1)\int_b^{\infty}e^{-y^2}\left[f^2(y)+f^2(x)\right]dy
$$
for fixed $x$ and $b > x$.
