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Let $X$ be a random variable with density $f(x)=|x|^{-3}1_{|x|\ge1}$ and $\phi_{X}(t)=E[e^{itX}]$. Show that $\forall t\in[-1,1] $ $$|\phi_{X}(t)-1-t^2log|t||\le3t^2$$ I noticed that $E[X]=0$, so $|\phi_{X}(t)-1-t^2log|t||=|E[e^{itX}-1-itX-t^2log|t|]|$. But $1,itX$ are the first two terms of the Taylor series of $e^{itX}$. Any ideas?

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  • $\begingroup$ It seems the following. $$\phi_{X}(t)=E[e^{itX}]=\int_{-\infty}^{+\infty} e^{itx}f(x)dx= \int_{-\infty}^{+\infty} e^{itx}f(x)dx=\int_{-\infty}^{+\infty} \frac{e^{itx}}{|x|^3}1_{|x|\ge1}dx=$$ $$2\int_{1}^{+\infty}\frac{e^{itx}+e^{-itx}}{x^3}dx=4\int_{1}^{+\infty}\frac{cos (xt)}{x^3}dx,$$ and the last integral can be estimated by different methods. $\endgroup$ Jun 7, 2015 at 16:17
  • $\begingroup$ For instance, by the Laplace method, see also this example. $\endgroup$ Jun 7, 2015 at 16:24
  • $\begingroup$ can you help me a little more? $\endgroup$ Jun 7, 2015 at 18:06
  • $\begingroup$ I am not a specialist in Laplace’s method too, :-( so let’s hope that an analyst will come about and point a simple way to estimate the integral. :-) $\endgroup$ Jun 7, 2015 at 18:38

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