Lower bound for integrals like $\int_1^{t+1}e^{-\sqrt{s}}s^{-1}ds$ Let
$$I(t) = \int_{1}^{t+1}\exp\left\{-c\frac{s^{1-\beta}}{1-\beta}\right\}s^{-2\beta}ds,$$
where $c$ is some positive constant and $\beta\in(0, 1)$.
Since the integral $I(t)$ given above could not be given in closed form, I want a lower bound of $I(t)$.
What is a lower bound for $I(t)$ which is increasing in $t$?
 A: Let $b:=\beta$. By substitution
\begin{equation}
    x=c\frac{s^{1-b}}{1-b},
\end{equation}
we have
\begin{equation}
    I(t)=kJ\Big(c\frac{(t+1)^{1-b}}{1-b}\Big),
\end{equation}
where
\begin{equation}
    k:=\Big(\frac{1-b}c\Big)^{\frac{1-2b}{1-b}},
\end{equation}
\begin{equation}
    J(y):=\int_{y_0}^y e^{-x}x^{-B}\,dx,\quad y_0:=\frac c{1-b}>0,\quad B:=\frac{2b}{1-b}>0. 
\end{equation}
So, it remains to provide a lower bound on $J(y)$ which is increasing in $y\ge y_0$. Using the inequality $x^{-B}\ge(e/y_0)^B\,e^{-Bx/y_0}$ for $x>0$, we have
\begin{equation}
\begin{aligned}
    J(y)&\ge \int_{y_0}^y e^{-x}(e/y_0)^B\,e^{-Bx/y_0}\,dx \\ 
    &=\Big(\frac e{y_0}\Big)^B\,\frac{y_0}{y_0+B}\,(e^{-y_0-B}-e^{-y-By/y_0}) =: H(y) 
\end{aligned}
\end{equation}
for $y\ge y_0$. Thus, we do have the lower bound $H(y)$ on $J(y)$ which is increasing in $y\ge y_0$. Also, $H(y_0)=0=J(y_0)$.
A: $\newcommand\b{\beta}$
Here is how I would write Iosif's answer, using the inequality $\ln u < u-1$.
Let
\begin{align}
u&=s^{1-\b}\\
du&=(1-\b)s^{-\b}ds\\
\frac{u^{-\b/(1-\b)}du}{1-\b}&=s^{-2\b}ds
\end{align}
So
\begin{align}
I(t)
&= \int_{1}^{t+1}\exp\left[-c\frac{s^{1-\beta}}{1-\beta}\right]s^{-2\beta}ds\\[1.5ex]
&= \int_{1}^{(1+t)^{1-b}}\exp\left[\frac{-cu}{1-\beta}\right]\frac{u^{-\b/(1-\b)}du}{1-\b}\\[1.5ex]
&= \frac{1}{1-\b}\int_{1}^{(1+t)^{1-\b}}\exp\left[\frac{-cu-\b\ln u}{1-\beta}\right]du\\[1.5ex]
&\ge \frac{1}{1-\b}\int_{1}^{(1+t)^{1-\b}}\exp\left[\frac{-cu-\b(u-1)}{1-\beta}\right]du\\[1.5ex]
&=\frac{e^{\b/(1-\beta)}}{-(\b+c)}\,\large{e^{au}\Big|_1^{(1+t)^{1-\b}}}\\[1.5ex]
&=\frac{e^{\b/(1-\beta)}}{\b+c}\large{\left(
e^{a}-
e^{a(1+t)^{1-\b}}
\right)}\\[1.5ex]
\end{align}
where $a=-(\b+c)/(1-\b)$.
This bound agrees with $I(t)$ when $t=0$, and is indeed increasing with $t$.
