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.

$$J(t) = \int^{\infty}_{0}{\exp\left(-t\left(x + \dfrac{4}{x+1} \right) \right)}\, dx$$

Solutions Collecting From Web of "Asymptotic expansion of $J(t) = \int^{\infty}_{0}{\exp(-t(x + 4/(x+1)))}\, dx$"

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 $t \to \infty$.

Please note that the leading exponential terms of your integral $J(t)$ can be expressed in terms of standard “modified Bessel functions” whose asymptotic expansion is known.
Let $z = \frac{x+1}{2} = e^\theta$, we have

$$\begin{align}
J(t)
& = e^t \int_0^\infty e^{-2t (\frac{x+1}{2} + \frac{2}{x+1})} dx
= 2 e^t \int_{\frac12}^\infty e^{-2t (z + z^{-1})} dz\\
& = 2 e^t \left[ \left( \int_0^\infty – \int_0^{\frac12} \right) e^{-2t (z + z^{-1})} dz \right]\\
& = 2 e^t \left[ \int_0^\infty e^{-2t (z + z^{-1})} dz – O\left( \int_0^{\frac12}e^{-2t(\frac12 + 2)} dz\right)\right]\\
& = 2e^t\int_{-\infty}^\infty e^{-4t\cosh\theta} e^\theta d\theta – O( e^{-4t} )\\
& = 4e^t\int_0^\infty e^{-4t\cosh\theta} \cosh\theta d\theta – O(e^{-4t})\\
& = 4e^t K_1(4t) – O(e^{-4t})
\end{align}
$$
where $K_\alpha(z)$ is the modified Bessel function
of the $2^{nd}$ kind. Using following asymptotic expansion of $K_\alpha(z)$:

$$K_\alpha(z) \sim \sqrt{\frac{\pi}{2z}} e^{-z} \left( 1 + \sum_{m=1}^{\infty} \frac{\prod_{k=1}^m (4\alpha^2 – (2k-1)^2)}{m!(8z)^m}\right)$$

We get
$$\begin{align}
J(t) \sim & \sqrt{\frac{2\pi}{t}} e^{-3t} \left(1 + \sum_{m=1}^{\infty} \frac{\prod_{k=1}^m (4 – (2k-1)^2)}{m!(8z)^m}\right) – O(e^{-4t})\\
= & \sqrt{\frac{2\pi}{t}} e^{-3t} \left( 1 +
\frac{3}{32t}-\frac{15}{2048 t^2}+\frac{105}{65536 t^3}-\frac{4725}{8388608 t^4}
+ \cdots \right) – O(e^{-4t})
\end{align}$$

Update

There are several ways to compute the asymptotic expansion of $J(t)$.
Other answers has shown how to cast it to an integral of the form
$$J(t) = \int e^{-t\varphi(y)} dy$$
and then expand $\varphi(y)$ to get desired result. One can also rewrite $J(t)$
to the form:
$$J(t) = \int_0^\infty e^{-ty}\psi(y) dy$$
and uses Watson’s Lemma
to obtain the asymptotic expansion.

Let I use $K_1(z)$ as an example.
In the integral representation of $K_1(z)$, let $1 + \eta = \cosh\theta$, we have

$$\begin{align}
K_1(z) & = \int_0^\infty e^{-z\cosh\theta} \cosh\theta d\theta\\
& = \int_0^\infty e^{-z(1+y)} \frac{1+y}{\sqrt{(y+1)^2-1}} dy\\
& = \frac{e^{-z}}{\sqrt{2}} \int_0^\infty e^{-zy}\frac{1+y}{\sqrt{y(1 + \frac{y}{2}})} dy\\
& \stackrel{formal}{=} \frac{e^{-z}}{\sqrt{2}} \int_0^\infty e^{-zy}\frac{1+y}{\sqrt{y}}
\left[ 1 + \sum_{m=1}^\infty \frac{(-1)^m}{m!} \left(\frac12 \right)_m \left(\frac{y}{2} \right)^m \right] dy\\
& = \frac{e^{-z}}{\sqrt{2}} \int_0^\infty \frac{e^{-zy}}{\sqrt{y}}
\left[ 1 + \sum_{m=1}^\infty \frac{(-1)^m}{2^m m!}
\left(\left(\frac12\right)_m – 2m\left(\frac12\right)_{m-1}\right)
y^m \right] dy\\
& = \frac{e^{-z}}{\sqrt{2}} \int_0^\infty \frac{e^{-zy}}{\sqrt{y}}
\left[ 1 + \sum_{m=1}^\infty \frac{(-1)^{m-1}}{2^m m!}
\left(\left(\frac12\right)_{m-1} (\frac12+m) \right)
y^m \right] dy\\
\end{align}$$
where $(\alpha)_m = \alpha(\alpha+1)\cdots(\alpha+m-1)$ is the rising
Pochhammer symbol.

In above transformations of the integral, the steps after the $\stackrel{formal}{=}$
are formal manipulations of equations. This is because the power series in
the integrand has a finite radius of convergence. However, this doesn’t stop
the last expression we arrived from being useful. Watson’s Lemma tell us if
the factor inside the $\left[ \cdots \right]$ doesn’t grow too fast, then we
can formally integrate the expression term by term and read off the asymptotic expansion of the integral directly. In this case, we have:
$$\begin{align}
K_1(z)
& \sim
\frac{e^{-z}}{\sqrt{2}} \frac{\Gamma(\frac12)}{\sqrt{z}}
\left[ 1 + \sum_{m=1}^\infty \frac{(-1)^{m-1}}{2^m m!}
\left(\left(\frac12\right)_{m-1} (\frac12+m) \frac{\Gamma(\frac12+m)}{\Gamma(\frac12)} \right)
\frac{1}{z^m} \right]\\
& = \sqrt{\frac{\pi}{2z}} e^{-z}\left[
1 + \sum_{m=1}^\infty \frac{(-1)^{m-1}}{2^m m!}
\left(\frac12\right)_{m-1} \left(\frac12\right)_{m+1} \frac{1}{z^m}
\right]\\
& = \sqrt{\frac{\pi}{2z}} e^{-z}\left[
1 + \sum_{m=1}^\infty \frac{\prod_{k=1}^m (4 – (2k-1)^2)}{8^m m!}
\frac{1}{z^m}
\right]
\end{align}
$$
If one really want, one can compute the asymptotic expansion of the $O(e^{-4t})$ term in $J(t)$ using same method.

This is solvable by Laplace’s method or the saddle point approximation.

If $t$ becomes large only those parts of the integrand where $f(x) = x + \frac{4}{x+1}$ is small will have a relevant contribution.

So one searches the minima of $f(x)$. In this case there is only one: $x_0 = 1$. One now approximates $f(x)$ by its Taylor expansion up to second order around $x_0$. The first order term will vanish since this is a minimum and the second order term will be positive, otherwise $x_0$ would not be a minimum but a maximum. In this case we arrive at:

$$f(x) = 3 + \frac{1}{2}(x-1)^2 + \mathcal{O}(x^3)$$

and with that, ignoring the higher order terms:

$$
J(t) = \int\limits_{0}^\infty dx\; e^{-3t – \frac{t}{2}(x-1)^2} = e^{-3t}\int\limits_0^\infty dx\; e^{-\frac{1}{2}(x-1)^2} = \sqrt{\frac{2\pi}{t}} e^{-3t}
$$

If you want to refine the solution, you can consider the remainder in the Taylor expansion as a perturbation. Let $g(x) = f(x) – 3 – \frac{1}{2}(x-1)^2$ be the error we made previously. Then you can approximate $e^{-tg(x)}$ around $x_0$ as $1-tg(x)$, because for large $t$ its contribution will vanish (there is no zeros order term in $g(x)$).

So you have:
$$
J(t) = e^{-3t}\int\limits_0^\infty dx\; e^{-\frac{t}{2}(x-1)^2}\left(1-\frac{t}{6}g”'(1)(x-1)^3 – \frac{t}{24}g””(1)(x-1)^4 + \ldots\right)
$$

The integrals for every summand can be solved. However they are not that nice. You may notice that all terms with $x$ to an odd power will vanish because a term exponential in $t$ will remain. All even powers will reproduce $\sqrt{\frac{2\pi}{t^k}}$ with a constant factor for some $k$, resulting in the expression given by Raymond Manzoni.