In a paper I'm reading, they say
Let $\delta$, $C_0$, and $n$ be positive constants. Then
$$\frac{\int_{\delta}^{\infty} \sqrt{n u^{\alpha}} e^{-nu^2 / (2C_0)}\,du}{\lambda_0^n} < \infty$$ for any fixed $\lambda_0 \in (e^{-\delta^2/(2C_0)},1)$.
I don't see why this is obvious. Why is this true?
Considering$$I=\int\sqrt{n}\, u^{\alpha /2} \,e^{-\frac{n }{2 c}u^2}\,du$$ let $u=\sqrt{\frac {2c}n}t$ to make $$I=\frac{(2c)^{\frac{\alpha+2 }{4}}}{n^{\frac{\alpha }{4}} }\int t^{\alpha /2}\,e^{-t^2}\,dt$$
$$J=\int_\epsilon^\infty t^{\alpha /2}\,e^{-t^2}\,dt=\frac{1}{2} \Gamma \left(\frac{\alpha +2}{4},\epsilon ^2\right)$$
Back to $u$ and $\delta$, this gives $$K=\int_\delta^\infty \sqrt{n}\, u^{\alpha /2} \,e^{-\frac{n }{2 c}u^2}\,du=2^{\frac{\alpha -2}{4}} c^{\frac{\alpha+2 }{4}} n^{-\alpha /4}\, \Gamma \left(\frac{\alpha +2}{4},\frac{n \delta ^2}{2 c}\right)$$
$$\frac K{\lambda^n} =k\, \lambda^{-n}n^{-\alpha /4}\, \Gamma \left(\frac{\alpha +2}{4},\frac{n \delta ^2}{2 c}\right)\qquad \text{where}\qquad k=2^{\frac{\alpha -2}{4}} c^{\frac{\alpha+2 }{4}} $$ Expanding as series for large values of $n$, this gives $$\log(\lambda^{-n}\,K)=-\frac{n \left(2 c \log (\lambda )+\delta ^2\right)}{2 c}+\log \left(\frac{c\, \delta ^{\frac{\alpha-2 }{2}}}{\sqrt{n}}\right)+O\left(\frac 1n\right)$$
To have convergence, we then need that the first coefficient be positive that is to say $$\log (\lambda )+\frac{\delta ^2}{2c} >0$$