A confusing mistake in evaluating $\int_{-\infty}^{\infty}{\frac{e^{-x^2}}{x^2+a^2}dx}$

Question

Assume a>0.
Using real method: $$\int_{-\infty}^{\infty}{\frac{e^{-x^2}}{x^2+a^2}dx}=\frac{e^{a^2}\pi\operatorname{erfc}(a)}{a}$$ Using residue: Let $C_L:|\Re z|\leq R$ , $C_R:|z|=R, \Im z>0$, their positive union is $C$. $$\int_C{f(z)dz}=2\pi i\operatorname{Res}(f,ai)=\frac\pi ae^{a^2}$$ I can easily prove $\int{_{C_R}}{f(z)dz}\to0$, so $$\int_{-\infty}^{\infty}{\frac{e^{-x^2}}{x^2+a^2}dx}=\frac\pi{2a}e^{a^2}$$ Mathematica shows that the first one is correct. So where is the mistake?

Answer 1

So the problem here is in the line

I can easily prove $\int_{C_R} f(z) \, dz \to 0...$

because unfortunately, it's false. Write $z = Re^{it}$; then

$$|\exp (-z^2)| = |\exp(-R e^{2it})| = \exp \Re(Re^{2it}) = \exp\big(-R \cos(2t)\big).$$

Now $t$ varies between $0$ and $\pi$; when $t \approx \pi/2$, the exponential is not decreasing much. There is a little angular sector whose angle is of order $1/R$ on which we have a uniform lower bound on the exponential; this has length of order $R$, and so we have a non-vanishing error term.

Answer 2

The mistake is the following: "I can easily prove $\int_C f(z) dz \to 0$".

I don't think this is true, since $f(z) \to \infty$ very fast around the imaginary axis.