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I need to estimate the growth of $$ \int_{-2c}^0\frac{\exp(-x^2)}{x+c}dx $$ as $c\to\infty$. In particular I want to show that $$ \int_{-2c}^0\frac{\exp(-x^2)}{x+c}dx=\mathcal{O}\left(\frac{1}{c}\right) $$ How should I proceed? Do you have any hint?

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Due to singularity at $x=-c$, this must be considered in the sense of principal value. For the estimation, just write $\int_{-2c}^{0}=\int_{-2c}^{-c/2}+\int_{-c/2}^{0}$. The first is exponentially decaying with $c$ (not hard to check), the second is trivially bounded from above (with $\frac{2}{c}\int_{-\infty}^{0}\exp(-x^2)\,dx$) and, if desired, can be similarly bounded from below (with $\frac{1}{c}\int_{-c/2}^{0}\exp(-x^2)\,dx$).

To go more precise, this integral is $\displaystyle\frac{1}{c}\int_{0}^{c}\frac{e^{-t^2}-e^{-(2c-t)^2}}{1-t/c}\,dt\thicksim\frac{\sqrt\pi}{2c}$.

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  • $\begingroup$ Wow, thank you! In the last expression everything gets pretty clear. If I can ask you one last thing, how do you show that the first term (in your first answer) has exponential decay in $c$? I don't how I am supposed to deal with the singularity at $x=-c$. $\endgroup$ – asd Oct 23 '18 at 11:13
  • $\begingroup$ Well, that integral is $\exp\big(-(c/2)^2\big)$ times an integral of a function that (after rearrangements dictated by the principal value sense) appears to be $\mathcal{O}(c)$, so it is $\mathcal{O}\big(c^2\exp(-c^2/4)\big)$. The same is seen if you take the last integral with the lower limit of $c/2$. $\endgroup$ – metamorphy Oct 23 '18 at 11:37
  • $\begingroup$ I guess I have to keep in mind that I'm always working with Cauchy principal value integral. Thank you again! $\endgroup$ – asd Oct 23 '18 at 21:04

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