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Prove: $\int_{1}^{t} (t-s)^{-\alpha} s^{-\beta}ds \leq Ct^{-\alpha}$, where $0<\alpha<1<\beta$ and $C$ is a constant, $t \geq 1$.

Progress: I already tried two ways to show it. One is divide $t^{-\alpha}$ on both sides and tried to approach Beta distribution however I failed since one of the parameters would be smaller than zero, rather than bigger than zero. Another way is after dividing like in the first method, I tried to bound the integral directly but it seems still not be able to be bounded. I appreciate any comments or solutions.

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Observe \begin{align} \int^t_1 (t-s)^{-\alpha}s^{-\beta}\ ds =&\ \int^t_{t/2+1/2} (t-s)^{-\alpha} s^{-\beta}\ ds + \int^{t/2+1/2}_{1} (t-s)^{-\alpha} s^{-\beta}\ ds\\ =:&\ I_1+I_2. \end{align} For $I_1$, we have \begin{align} I_1 \leq \frac{1}{(t/2+1/2)^\beta} \int^t_{t/2+1/2} \frac{ds}{(t-s)^{\alpha}} \leq C \frac{(t-1)^{1-\alpha}}{(t+1)^\beta} \leq C't^{1-\alpha-\beta} \end{align} when $t\gg 1$.

For $I_2$, we have \begin{align} I_2 \leq C\frac{1}{(t/2-1/2)^\alpha} \int^{t/2+1/2}_1 s^{-\beta}\ ds \leq C't^{-\alpha} \end{align} when $t\gg 1$.

Hence it follows \begin{align} I_1 + I_2 \leq C_1t^{1-\alpha-\beta}+C_2t^{-\alpha} \leq C_3t^{-\alpha}. \end{align}

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  • $\begingroup$ Could you be more specific about estimating $I_1$ and$I_2$, especially $C'$? $\endgroup$ – Topoguy Oct 25 '16 at 21:47
  • $\begingroup$ @JackD'Aurizio They don't depend on $t$. Observe for $I_1$, it's not hard to show that $C=1-\alpha$ since $(t-1)^{1-\alpha} \leq (t+1)^\beta$ for $t\gg 1$. And for $I_2$, we can see that $C = \int^\infty_1 \frac{ds}{s^\beta}$. $\endgroup$ – Jacky Chong Oct 25 '16 at 22:42
  • $\begingroup$ @JackyChong: I owe you some apologies, since in my observation I set $\beta=1$, that is not really allowed by the given constraints. Splitting in halves is perfectly fine. (+1) and sorry if I caused you an unnecessary time loss. $\endgroup$ – Jack D'Aurizio Oct 25 '16 at 23:01
  • $\begingroup$ @JackD'Aurizio No apology necessary. We are only trying to get things right. $\endgroup$ – Jacky Chong Oct 25 '16 at 23:05
  • $\begingroup$ @JackD'Aurizio Hi, where is $t^{1-\beta}$ in your answer? I mean you are trying to show $t^{1-\beta}\int_{1/t}^{1}(1-u)^{-\alpha}u^{-\beta}\,du\color{red}{\leq} C$, but I only see you just showed the bound for the integral. $\endgroup$ – Topoguy Oct 26 '16 at 0:21

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