Decay for Convolution Integral

Question

I am trying to find a bound for the decay of $$\int_0^t (1+s)^{-a} t^{-b\frac{t-s}{t}} ds,$$ with constants $a\geq 1$ and $b>0$. Ideally it should look like $$\int_0^t (1+s)^{-a} t^{-b\frac{t-s}{t}} ds\leq c (1+t)^{-f(a,b)}$$ for at least $t>>1$.

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From Mathematica it seems that the decay above is true. Also $$\int_0^t (1+s)^{-a} t^{-b\frac{t-s}{t}} ds = (1+t)^a E_a\left(-b\frac{\ln t}{t}\right)-(1+t)E_a\left(-b(1+t)\frac{\ln t}{t}\right),$$ where $E_n(z)$ is the exponential integral function and this could be converted into $$t^{-b(1+1/t)}\left(-b\frac{\ln t}{t}\right)^{a-1}\left(\Gamma\left(1-a,-b\frac{\ln t}{t}\right)-\Gamma\left(1-a,-b(1+t)\frac{\ln t}{t}\right) \right),$$ but I still don't know how to estimate the incomplete Gamma function difference.

Answer 1

I even found a better bound using a similar approach, but only for the case $\beta<\alpha$.

For $\alpha, \beta >0$, $\beta < \alpha$ one has \begin{align*} \int_0^t (1+s)^{-1-\alpha} t^{-\beta\frac{t-s}{t}}\ ds \leq c t^{-\beta}. \end{align*}

Let $t\geq 1$, chose $\delta$ such that $0<\frac{\beta}{\alpha}<\delta<1$ and split the integral into \begin{align*} \int_0^t (1+s)^{-1-\alpha} t^{-\beta\frac{t-s}{t}}\ ds &= t^{-\beta} \int_0^t (1+s)^{-1-\alpha} t^{\beta\frac{s}{t}}\ ds \\ &=t^{-\beta} \left(\int_0^{t^\delta} (1+s)^{-1-\alpha} t^{\beta\frac{s}{t}}\ ds+\int_{t^\delta}^t (1+s)^{-1-\alpha} t^{\beta\frac{s}{t}}\ ds \right) \end{align*} and show that both of these integrals are bounded. First \begin{align*} \int_0^{t^\delta} (1+s)^{-1-\alpha} t^{\beta\frac{s}{t}}\ ds &\leq \int_0^{t^\delta} (1+s)^{-1-\alpha} t^{\beta t^{\delta-1}}\ ds\\ &=e^{\beta \frac{\ln t}{t^{1-\delta}}}\int_0^{t^\delta} (1+s)^{-1-\alpha} \ ds\\ &=c(\beta,\delta) \frac{1}{-\alpha} \left((1+t^\delta)^{-\alpha}-1\right)\\ &\leq c(\alpha,\beta,\delta) \end{align*} and for the second one \begin{align*} \int_{t^\delta}^t (1+s)^{-1-\alpha} t^{\beta\frac{s}{t}}\ ds &\leq \int_{t^\delta}^t (1+s)^{-1-\alpha+\frac{\beta}{\delta}} s^{-\frac{\beta}{\delta}}t^{\beta\frac{s}{t}}\ ds \\ &\leq \int_{t^\delta}^t (1+s)^{-1-\alpha+\frac{\beta}{\delta}} (t^\delta)^{-\frac{\beta}{\delta}}t^{\beta\frac{s}{t}}\ ds \\ &= \int_{t^\delta}^t (1+s)^{-1-\alpha+\frac{\beta}{\delta}} t^{\beta(\frac{s}{t}-1)}\ ds \\ &\leq \int_{t^\delta}^t (1+s)^{-1-\alpha+\frac{\beta}{\delta}}\ ds \\ &=\frac{1}{\alpha-\frac{\beta}{\delta}} \left((1+t^\delta)^{-\alpha+\frac{\beta}{\delta}}-(1+t)^{-\alpha+\frac{\beta}{\delta}}\right)\\ &\leq c(\alpha,\beta,\delta) \end{align*} as $\alpha> \frac{\beta}{\delta}$. For $t \leq 1$ trivially \begin{align*} \int_0^t (1+s)^{-1-\alpha} t^{-\beta\frac{t-s}{t}}\ ds &\leq t^{-\beta}\ \int_0^1 1\ ds\\ &= t^{-\beta}. \end{align*}

Answer 2

I found a way to get a decay but it is not really sharp. Nonetheless for those who it might help in the future here it is:

For $t\gg 1$, $\alpha, \beta >0$, $\frac{\beta}{1+\alpha}<1$ \begin{align*} \int_0^t (1+s)^{-1-\alpha} t^{-\beta\frac{t-s}{t}}\ ds = \mathcal{O}\left(t^{-\beta\frac{\alpha}{1+\alpha}}\right) \end{align*} (Note that I switched $a\to \alpha + 1$)

Proof:
Split the integral at $0<\epsilon<t$ such that the first part can be estimated by \begin{align*} \int_0^\epsilon (1+s)^{-1-\alpha} t^{-\beta\frac{t-s}{t}}\ ds&\leq t^{-\beta}\int_0^\epsilon e^{\beta \frac{\ln t}{t}s}\ ds\\ &= \frac{t^{1-\beta}}{\beta \ln t} \left(e^{\beta\frac{\ln t}{t}\epsilon}-1\right)\\ &= \epsilon t^{-\beta} +\epsilon t^{-\beta} \sum_{k=2}^\infty \frac{\left(\beta\frac{\ln t}{t}\epsilon\right)^{k-1}}{k!}\\ &\leq c\epsilon t^{-\beta} \end{align*} and the second part by \begin{align*} \int_\epsilon^t (1+s)^{-1-\alpha} t^{-\beta\frac{t-s}{t}}\ ds &\leq \int_\epsilon^t (1+s)^{-1-\alpha} \ ds\\ &= \frac{1}{\alpha} \left(\frac{\epsilon^{-\alpha}}{\left(1+\frac{1}{\epsilon}\right)^{\alpha}}-\frac{t^{-\alpha}}{\left(1+\frac{1}{t}\right)^{\alpha}}\right). \end{align*} So for $\epsilon = t^\delta$ with $0<\delta < 1$ the total integral is bounded by \begin{align*} \int_0^t (1+s)^{-1-\alpha} t^{-\beta\frac{t-s}{t}}\ ds &\leq \mathcal{O}\left(t^{\delta-\beta}\right)+\mathcal{O}\left(t^{-\alpha \delta}\right)+\mathcal{O}\left(t^{-\alpha}\right)\\ &\leq \mathcal{O}\left(t^{\delta-\beta}\right)+\mathcal{O}\left(t^{-\alpha \delta}\right) \end{align*}

Optimizing $\delta$ gives $\delta = \frac{\beta}{1+\alpha}$ and the therefore claim.