Verify the estimate $\int_{0}^v x^B \mathrm{e}^{\beta x^s} {\rm d}x = \frac{1}{\beta s} v^{B+1-s} \mathrm{e}^{\beta v^s} (1 + o(1))$, $v \to +\infty$ The authors   Butucea and Tsybakov in the paper  Sharp optimality for density deconvolution with
dominating bias (arXiv pdf link) claim the following assertion (see Lemma 6, p.29): For any $B\in \mathbb{R}$ and for any $\beta>0$, $s>0$,
$$\int_{0}^v x^B \mathrm{e}^{\beta x^s} {\rm d}x = \frac{1}{\beta s} v^{B+1-s} \mathrm{e}^{\beta v^s} (1 + o(1)), \quad v \to +\infty.$$
I need more detailed arguments for the proof of this assertion. Can someone help me?
 A: If $B\le -1$ then the integral is infinite  due to singularity at $x=0$. Otherwise, we proceed as follows:

*

*Special case $B=s-1$ (e.g. $s=1,B=0$). Then the integral is computable in closed form, $$\int_{0}^v x^{s-1} e^{\beta x^s} dx = \frac1{\beta s}\int_{0}^v d(e^{\beta x^s})=\frac1{\beta s} (e^{\beta v^s}-1) = \frac1{\beta s} e^{\beta v^s}(1+ o(1)). $$

*Special case $s=1, B>0$. Integrate by parts like so:

\begin{align}\int_{0}^v \underbrace{x^{B}}_{=u} \underbrace{e^{\beta x}}_{=v'} dx 
&= \frac1{\beta}v^B e^{\beta v} -\frac{B}\beta \int_0^vx^{B-1} e^{\beta x} dx \\
&= \frac1\beta v^Be^{\beta v}\Big(1-Be^{-\beta v}\int_0^v (x/v)^{B-1} e^{\beta x} d(x/v)\Big)
\\
&= \frac1\beta v^Be^{\beta v}\Big(1-B\underbrace{\int_0^1 y^{B-1} e^{\beta  v(y-1)} dy}_{\to 0 \text{ by DCT}}\Big) = \frac1{\beta\cdot 1} v^Be^{\beta {v^1}}(1+o(1)).
\end{align}


*For $s=1$,  $B\in(-1,0)$, we cannot precisely proceed as above because the integration by parts doesn't work at the endpoint $x=0$. But we can simply split $\int_0^v$ into $\int_0^1 + \int_1^v$, and treat the $\int_0^1$ piece as an $O(1)$ error. Now integration by parts and DCT again gives
$$\int_{0}^v\!\!\! x^{B} e^{\beta x} dx=O(1) +\frac1\beta v^B e^{\beta v} - \frac{e^\beta}\beta - \frac {Bv^Be^{\beta v}}\beta \int_{\frac1v}^1\! y^{B-1} e^{\beta v(y-1) } dy = \frac1{\beta\cdot1} v^Be^{\beta v^1}(1+o(1)).$$


*General case.  We set $X=x^s, V=v^s$. then $dX=s x^{s-1}dx = X^{1-1/s} dx$, and the integral transforms to
$$ \int_{x=0}^v x^{B} e^{\beta x^s} dx 
 =\frac1s\int_{X=0}^{V} X^{(B+1)/s-1} e^{\beta X}dX.$$
Cases 1, 2, and 3 lead to the answer
$$ \int_{x=0}^v x^{B} e^{\beta x^s} dx =\frac1{\beta s} V^{(B+1)/s-1} e^{\beta V}(1+o_{V\to\infty}(1)) = \frac1{\beta s} v^{B+1-s} e^{\beta v^s}(1+o_{v\to\infty}(1)). $$
A: Assume that $B>-1$, and $\beta,s>0$. Then, using L'Hôpital's rule, we obtain
\begin{align*}
\mathop {\lim }\limits_{v \to  + \infty } \frac{{\int_0^v {x^B \mathrm{e}^{\beta x^s } \mathrm{d}x} }}{{v^{B + 1 - s} \mathrm{e}^{\beta v^s } }}& = \mathop {\lim }\limits_{v \to  + \infty } \frac{{v^B \mathrm{e}^{\beta v^s } }}{{(B + 1 - s)v^{B - s} \mathrm{e}^{\beta v^s }  + \beta sv^B \mathrm{e}^{\beta v^s } }} \\ & = \mathop {\lim }\limits_{v \to  + \infty } \frac{1}{{(B + 1 - s)v^{ - s}  + \beta s}} = \frac{1}{{\beta s}}.
\end{align*}
A: Here's an initial approach :
$$
\int_0^v x^B e^{\beta x^s}dx=\frac{1}{\beta s}\int_0^v x^{B+1-s}\beta s x^{s-1}e^{\beta x^s}dx
$$
Let $f=x^{B+1-s}$, then $f'=(B+1-s)x^{B-s}$.
Let $g'=\beta s x^{s-1}e^{\beta x^s}$, then $g=e^{\beta x^s}$.
By assuming $B+1-s\geq0$, we get
\begin{align*}
\frac{1}{\beta s}\int_0^v x^{B+1-s}\beta s x^{s-1}e^{\beta x^s}dx&=\Bigg[\frac{1}{\beta s}x^{B+1-s}e^{\beta x^s}\Bigg]_0^v - \frac{1}{\beta s}\int_0^v (B+1-s)x^{B-s}e^{\beta x^s}dx\\
&=\frac{1}{\beta s}v^{B+1-s}e^{\beta v^s}-\frac{1}{\beta s}\int_0^v (B+1-s)x^{B-s}e^{\beta x^s}dx.\\
\end{align*}
I have fewer ideas on how to go with further calculations.
Hope it helps a bit.
