Laplace method for $\int_0^1 dx x^\gamma \frac{\partial}{\partial x} P\big(\frac{u}{x}\big) $ where $\gamma \gg 1$ and $P(\infty)\rightarrow 0$ I have a structure like
$$ F(u) = \int_0^1dx x^\gamma \frac{\partial}{\partial x}P\big(\frac{u}{x}\big) $$
where $\gamma\gg 1$ is a positive (noninteger) exponent and $P$ is an unknown function with $P(\pm \infty) = 0$. Since the integrand is dominated by the value $x=1$, I wonder if I can perform a Laplace-type asymptotic expansion of this integral despite not knowing anything about $P$.
Naively, I want to say something like $x^\gamma \approx 1$ so that I can trivially integrate for $F(u) = P(u) - P(\infty) = P(u)$, but I'm not sure this handwavy approach is correct and I'm unclear how to proceed. I am aware of some similar problems, but these do not involve unknown functions (like $\partial P(u/x)/\partial x$). I would be very happy to read any thoughts on if such a Laplace-like integral can be done with unspecified $P$.
 A: Though the major contribution to the integral comes from the region $x\simeq1$, this region becomes very narrow as $\gamma$ increases. The naive trick cannot apply.
To simplify the problem, an integration by part can be  performed first:
\begin{align}
 F(u) &= \int_0^1 x^\gamma \frac{\partial}{\partial x}P\left(\frac{u}{x}\right)\,dx\\
 &=P(u)-\gamma\int_0^1 x^{\gamma-1} P\left(\frac{u}{x}\right)\,dx
\end{align}
Now, by substituting $x=\exp(-t)$, the integral is transformed into a Laplace integral:
\begin{equation}
 F(u)=P(u)-\gamma\int_0^\infty e^{-\gamma t}P\left( ue^t \right)\,dt
\end{equation}
If, for $s\to 0^+$ the function $P(ue^t)$ can be expanded as
\begin{equation}
 P(ue^t)\sim\sum_{s=0}^{\infty}a_{s}t^{(s+\lambda-\mu)/\mu}
\end{equation}
where $\lambda$ and $\mu$ are positive constants, Watson lemma gives the asymptotic expansion
\begin{equation}
 \int_{0}^{\infty}e^{-\gamma t}q(t)\mathrm{d}t\sim\sum_{s=0}^{\infty}\Gamma\left(%
\frac{s+\lambda}{\mu}\right)\frac{a_{s}}{\gamma^{(s+\lambda)/\mu}}
\end{equation}
In particular, if $P(y)$ is analytical at $y=u$, then $\lambda=\mu=1$ and
\begin{equation}
 P(ue^{t})=P(u)+suP'(u)+\frac{s^2}{2}u\left( P'(u)+uP''(u) \right)+\cdots
\end{equation}
Thus
\begin{align}
 F(u)&=P(u)-\gamma\left( \frac{P(u)}{\gamma}+2\frac{uP'(u)}{\gamma^2} +3u\frac{ P'(u)+uP''(u)}{\gamma^2}+\cdots\right)\\
 &= -\frac{2uP'(u)}{\gamma}-3\frac{u\left( P'(u)+uP''(u) \right)}{\gamma^2}+\cdots
\end{align}
