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Let $T$ a positive number, $s\in\mathbb{R}$ and $s-T\leq\sigma\leq s+T$. I need to prove the following inequality $$\int_{s-T}^{s+T}\frac{1}{\sqrt{1+A^2(t-\sigma)^2}}dt\leq C\frac{1}{\sqrt{1+A^2}}\,,$$ with $A$ positive. I tried with the substitution $A(t-\sigma)=t'$, but I don't get the desired result. Any suggestion? The integral in the left hand side can be computed and the result is $$\frac{1}{A}[\sinh^{-1}(A(s+T-\sigma))-\sinh^{-1}(A(s-T-\sigma))]\,.$$

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    this integral is easily done analitically – 2017-02-01
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    @tired It's what I've done with the substitution I mentioned above, but I didn't get the desired estimate. – 2017-02-01
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    @tired The result is $\frac{1}{A}[\sinh^{-1}(A(s+T-\sigma))-\sinh^{-1}(A(s-T-\sigma))]$. – 2017-02-01
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    If you could show that the left hand side is maximized for $\sigma = s$, then you could bound it by $$\int_{s-T}^{s+T} \frac{1}{\sqrt{1+A^2(t-s)^2}} \,dt$$ which after a substitution of $u = t - s$ becomes $$\int_{-T}^{T} \frac{1}{\sqrt{1+A^2u^2}} \,du$$, which gets rid of the $\sigma$s and $s$s. Then handle the cases where $|u| \leq 1$ and $|u| > 1$ separately, since the latter integrand can be bounded by something like the RHS of your inequality. – 2017-02-01
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    @CodeLabMaster If you want to get rid of the $\sigma$s and the $s$s, just by using the fact that $\sigma,t\in[s-T,s+T]$, you can bound the integral in the LHS with $$\int_{-2T}^{2T}\frac{1}{\sqrt{1+A^2u^2}}\,.$$ – 2017-02-02

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