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Find the limit $$ \large \space\lim\limits_{n\to\infty}\sqrt{n}\int\limits_{-\infty}^{+\infty}\frac{\cos t}{\left(1+t^2\right)^n}dt $$

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  • $\begingroup$ Why is the limit clear? @AntonioVargas What is this called? $\endgroup$ – Permian Dec 7 '15 at 20:30
  • $\begingroup$ Now I remember. $\endgroup$ – Permian Dec 7 '15 at 20:31
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    $\begingroup$ Rewrite as $$2 \sqrt{n} \int_0^{\infty} dt \, \cos{t} \, e^{-n \log{(1+t^2)}}$$ and use Laplace's method. $\endgroup$ – Ron Gordon Dec 7 '15 at 21:05
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For big $n$ the integrand is cleary dominated from the region around $t=0$. we therefore might write ($\epsilon<<1$ ) $$ I(n)\approx \int_{-\epsilon}^{\epsilon}\cos(t)\frac{1}{(1+t^2)^n}=\int_{-\epsilon}^{\epsilon}\cos(t)e^{-n\log(1+t^2)}\approx\int_{-\epsilon}^{\epsilon}\cos(t)e^{-nt^2}\approx\int_{-\epsilon}^{\epsilon}e^{-nt^2} $$

where we have used $\log(1+x)\approx x$ and $\cos(x)\approx 1$ for $x<<1$. The usual trick is now that we can extend the limits of integration back to $\pm \infty$ inducing only an exponentially small error (All steps can be made rigourous by the method of steepest descent). We are therefore left with a standard Gaussian integral

$$ I(n)\approx\int_{-\infty}^{\infty}e^{-nt^2}=\frac{\sqrt{\pi}}{\sqrt{n}} $$

and we can conclude that

$$ \lim_{n\rightarrow \infty} \sqrt{n}I(n)= \lim_{n\rightarrow \infty} \sqrt{n}\frac{\sqrt{\pi}}{\sqrt{n}}=\sqrt{\pi} $$

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  • $\begingroup$ "All steps can be made rigourous by the method of steepest descent" What does this mean? Can you provide me to a relevant link that explains this? THANKYOU $\endgroup$ – GohP.iHan Dec 10 '15 at 3:33
  • $\begingroup$ @GohP.iHan my approach is pretty heuristic. For example the statement that one can reexpand the intervall of integration. This can be made rigorous by using Steepest Descent method for example $\endgroup$ – tired Dec 10 '15 at 17:47
  • $\begingroup$ Oh just this one? Got it, let me learn something new!! TQ $\endgroup$ – GohP.iHan Dec 10 '15 at 18:35

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