# Asymptotic growth of $\int_{-2c}^0\frac{\exp(-x^2)}{x+c}dx$

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?

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 $$\displaystyle\frac{1}{c}\int_{0}^{c}\frac{e^{-t^2}-e^{-(2c-t)^2}}{1-t/c}\,dt\thicksim\frac{\sqrt\pi}{2c}$$.
• 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 $x=-c$. – asd Oct 23 '18 at 11:13
• 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 $c/2$. – metamorphy Oct 23 '18 at 11:37