Asymptotic expansion of $J(t) = \int^{\infty}_{0}{\exp(-t(x + 4/(x+1)))}\, dx$ I want to derive an asymptotic expansion for the following Bessel function. I think I need to rewrite it in another form, from which I can integrate it by parts. I am interested in obtaining the expansion up to second order as $t$ approaches infinity.
$$J(t) = \int^{\infty}_{0}{\exp\left(-t\left(x + \dfrac{4}{x+1} \right) \right)}\, dx$$

First make the (cosmetic) change of variables $x = y+1$, so that
$$
\begin{align}
J(t) &= \int^{\infty}_{0} \exp\left[-t\left(x + \dfrac{4}{x+1} \right) \right]\, dx \\
&= \int_{-1}^{\infty} \exp\left[-t\left(y + 1 + \dfrac{4}{y+2} \right) \right]\, dy \\
&= \int_{-1}^{\infty} \exp\Bigl[-t f(y) \Bigr]\,dy.
\end{align}
$$
Now $f(y)$ has a minimum at $y=0$, and near there we have
$$
f(y) = 3 + \frac{y^2}{2} - \frac{y^3}{4} + \frac{y^4}{8} + \cdots.
$$
We would like to introduce a new variable by
$$
3 + \frac{z^2}{2} = 3 + \frac{y^2}{2} - \frac{y^3}{4} + \frac{y^4}{8} + \cdots
$$
or, taking the principal branch of the square root,
$$
z = y \sqrt{1 - \frac{y}{2} + \frac{y^2}{4} + \cdots},
$$
which of course would only hold in a small neighborhood of the origin.  In general we can solve for $y$ in this equation using series reversion (the Lagrange formula, say), but in this case we can write down the answer explicitly as
$$
\begin{align}
y &= \frac{1}{4} \left(z^2 + z \sqrt{16+z^2}\right) \\
&= z + \frac{z^2}{4} + \frac{z^3}{32} - \frac{z^5}{2048} + O\left(z^7\right).
\end{align}
$$
Thus we have
$$
dy = \left[1+\frac{z}{2}+\frac{3 z^2}{32}-\frac{5 z^4}{2048} + O\left(z^6\right)\right]dz.
$$
The integrals over the tails of the intervals are exponentially small as $t \to \infty$, and by following the usual steps in the Laplace method we arrive at the expression
$$
J(t) = \int_{-\infty}^{\infty} \exp\left[-t \left(3 + \frac{z^2}{2}\right)\right] \left(1+\frac{z}{2}+\frac{3 z^2}{32}-\frac{5 z^4}{2048}\right)\,dz + O\left(\int_{-\infty}^{\infty} \exp\left[-t \left(3 + \frac{z^2}{2}\right)\right] z^6\,dz\right).
$$
We can integrate term-by-term and conclude that
$$
J(t) = e^{-3t} \sqrt{\frac{2\pi}{t}} \left[1 + \frac{3}{32} t^{-1} - \frac{15}{2048} t^{-2} + O\left(t^{-3}\right)\right]
$$
as $t \to \infty$.
