Compute the main order asymptotics of the integral $\int_0^{\infty}e^{itx-e^x}dx$, as $\mid t \mid \to \infty$. I have found this integral interesting because it does not fall into any of the Fourier/steepest descent/integration by parts methods to compute asymptotics! The Mellin transform of this function is also ill-defined !
Has someone come across a similar oscillating integral ? Any novel ideas would be greatly appreciated.
Thanks in advance.
 A: OP's integral is an incomplete Gamma function
$$I(t)~:=~\int_{\mathbb{R}_+} \! \mathrm{d}z~e^{it -\exp(z)}
~\stackrel{u=\exp(z)}{=}~ 
\int_1^{\infty}\! \frac{\mathrm{d}u}{u} u^{it}e^{-u}
~=~\Gamma(it,1),\qquad t\in \mathbb{R}\backslash\{0\}. \tag{1}$$
Here is one way to extract the asymptotic behaviour

$$I(t)~\sim~\frac{i}{et}\quad\text{for}\quad |t|\to \infty. \tag{2}$$

Deform the integration contour $\mathbb{R}_+$ to

*

*Vertically along the line $x=0$: From the origin $z=0$ to $z=i{\rm sgn}(t)\frac{\pi}{4}$.


*Horizontally along the line $y=\frac{\pi}{4}{\rm sgn}(t)$: From $z=\frac{i\pi}{4}{\rm sgn}(t)$ to $z=\infty+\frac{i\pi}{4}{\rm sgn}(t)$.


*It is easy to see that the third leg at infinity does not contribute.
The corresponding integral becomes
$$ J(t)~:=~ t I(t) ~=~J_1(t)+J_2(t),\qquad t\in \mathbb{R}\backslash\{0\},\tag{3}$$
where
$$J_1(t)~:=~i|t| \int_0^{\pi/4} \! \mathrm{d}y~e^{-|t|y-\exp(iy~{\rm sgn}(t))} 
~\stackrel{y^{\prime}=|t|y}{=}~
i \int_0^{|t|\pi/4} \! \mathrm{d}y^{\prime}~e^{-y^{\prime}-\exp(iy^{\prime}/t)}$$
$$\quad\longrightarrow\quad i \int_0^{\infty} \! \mathrm{d}y^{\prime}~e^{-y^{\prime}-\exp(0)}~=~\frac{i}{e}\quad\text{for}\quad |t|\to \infty, \tag{4} $$
and
$$J_2(t)~:=~|t|e^{-|t|\pi/4} \int_0^{\infty} \! \mathrm{d}x~e^{-itx-\exp(x)\exp[i~{\rm sgn}(t)~\pi/4]}$$
$$\quad\longrightarrow\quad 0\quad\text{for}\quad |t|\to \infty. \tag{5} $$
In eq. (4) we used Lebesgue's dominated convergence theorem with majorant function $$g(y^{\prime})~=~e^{-y^{\prime}}.\tag{6}$$
In eq. (5) we used the inequality
$$\left| \int\! \mathrm{d}x ~f(x) \right| ~\leq~ \int\! \mathrm{d}x ~|f(x)|.\tag{7} $$
A: The integral is doable by steepest descent, you just have to take the endpoints into account as well. If $t > 0$, then the direction of the steepest descent at $x = 0$ is $i$:
$$x = i \xi, \\
\int_0^\infty \exp(i t x - e^x) dx \sim \\
i \exp(-e^{i \xi}) \Big \rvert_{\xi = 0} \int_0^\infty e^{-t \xi} d\xi =
\frac i {e t}.$$
Other contributions are negligible because $\operatorname{Re}(i t x)$ is negative in the upper half-plane.
