Asymptotic expansion of $\int \frac{1}{(y^2+c^2)^n}\, e^{-\frac{\lambda}{2} y^2} dy$

Question

Let $\lambda>0$. Are there $c>0,K>0$ such that for all $n\in\mathbb N$ $$ \int_{-\infty}^{\infty} x^{2n}\, e^{-\frac{\lambda}{2} x^2} dx\,\ \cdot\ \int_{-\infty}^{\infty} \frac{1}{(y^2+c^2)^n}\, e^{-\frac{\lambda}{2} y^2} dy \ \,\leq\, K^n \ \ ?$$

The first integral by change of variable $x=\pm\sqrt{2t/\lambda}$ rewrites as

$$ \int_{-\infty}^{\infty} x^{2n}\, e^{-\frac{\lambda}{2} x^2}\,dx \,=\, 2\,\left(\frac{2}{\lambda}\right)^{n-\frac{1}{2}}\! \int_0^\infty t^{n-\frac{1}{2}}\,e^{-t}\,dt \,=\, 2\,\left(\frac{2}{\lambda}\right)^{n-\frac{1}{2}}\Gamma\Big(n+\frac{1}{2}\Big) \,=\\ =\, \frac{\sqrt{2\pi}}{\lambda^{n-\frac{1}{2}}}\,(2 n-1)!! $$

How can I find an asymptotic expansion of the second integral? Is there a term compensating for the $(2n-1)!!$ which grows faster than exponentially?

Answer 1

The answer is no. Laplace method can be used to compute asymptotics of the second integral:

$$ I_n\equiv \int_{-\infty}^\infty \frac{1}{(y^2+c^2)^n}\,e^{-\frac{\lambda}{2}y^2}\,dy \,=\, \int_{-\infty}^\infty g(y)\,e^{-nF(y)}\,dy$$ where: $F(y)\equiv\log(y^2+c^2)$ has a unique minimum point in $y=0$ with $F''(0) = 2/c^2\,$, and $g(y)\equiv e^{-\frac{\lambda}{2} y^2}\,$. Hence as $n\to\infty$

$$ I_n \,\sim\, \sqrt{\frac{2\pi}{n\,(2/c^2)}}\, g(0)\,e^{-n F(0)} \,=\,\frac{\sqrt{\pi}}{\sqrt{n}\,c^{2n-1}} \;.$$