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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?

Kashif
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  • The integral converges.How can the infinity arise? – user Jun 02 '20 at 06:24
  • Why does this integral converge? – Kashif Jun 02 '20 at 06:26
  • $\int_{\delta}^{\infty} u^{\beta} e^{-cu^{2}} du <\infty$ for any positive numbers $\beta $ and $c$. – Kavi Rama Murthy Jun 02 '20 at 06:26
  • Here's an informal argument for convergence: $e^{-u^2}$ decreases faster than any polynomial, even when multiplied by $u^\alpha$ for any finite $\alpha$ (you can see this by Taylor expanding). The so often called "p-test" tells you that $\int_0^\infty u^{-p} ,du$ converges if $ p > 1$, which is definitely the case by looking at the leading order term of the taylor series for $u^\alpha e^{-u^2}$. – AlkaKadri Jun 02 '20 at 06:47
  • @AlkaKadri I'm not following. If I do what you said, I get $\sum_{k=0}^{\infty} \frac{u^{\alpha - 2k}}{k!}$ as the Taylor expansion of $u^{\alpha} e^{-u^2}$. How does the integral of this converge when the first term is $u^{\alpha}$? – Kashif Jun 04 '20 at 20:09
  • Nvm, I got it. Noted that $e^{u^2} = \sum_{k=0}^{\infty} u^{2k}/k! > u^{2a+2}/(a+1)!$ for any $a$ and then $u^a e^{-u^2} < \frac{(a+1)!}{u^{a+2}}$, which converges upon integration. – Kashif Jun 05 '20 at 19:05

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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 $$\log (\lambda )+\frac{\delta ^2}{2c} >0$$

Claude Leibovici
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  • What does $\Gamma \left(\frac{\alpha +2}{4},\frac{n \delta ^2}{2 c}\right)$ mean?

    And I don't understand how you got the $-\frac{ n (2c \log(\lambda) + \delta^2) }{2c} + \log \left( \frac{ c \delta^{\frac{\alpha-2}{2}} }{\sqrt{n}} \right) + O(\frac{1}{n})$ either. Can you elaborate further?

     – Kashif Jun 04 '20 at 17:13
  • @Glassjawed. The incomplete gamma function. – Claude Leibovici Jun 05 '20 at 01:59