How do you do this integral? I'm trying to calculate the following integral (using special functions and/or contour integration):
$$I(x)=\int_{-\infty}^{\infty}e^{ix t}t\,e^{-i\sqrt{t^2+a^2}}
\,dt$$
Here, $x$ is a real number/variable, $a$ is a real number, and $i$ is the imaginary unit. I'm only really interested in the $x\rightarrow \infty$ ($x$ is a real number) limiting behavior of this function, but it would be especially nice if I could get some sort of closed form expression for this integral.
If you also know of some particular method I could use to calculate this integral, please tell me. In the form that I currently have it in, I am compelled to think that some sort of an asymptotic expansion may work. I suppose I could also change the bounds of integration, write $t=\exp (\ln(t))$, and try the method of steepest descents.
 A: *

*OP's integrand in his integral
$$I(x)~:=~\int_C\! \mathrm{d}w~we^{i\{xw -\sqrt{w^2+a^2}\}},$$ $$  w~\equiv~u+iv, \qquad u,v~\in~\mathbb{R},  \qquad x~>~1, \qquad  a~\geq~ 0,\tag{1}$$
is not ${\cal L}^1$-integrable if we chose the integration contour $C$ to be the real $u$-axis in the complex $w$-plane. To repair this, we slightly deform the integration contour $C$ into the upper half-plane 
$$u\quad\mapsto\quad w~=~u+i\underbrace{\epsilon|u|}_{=v}, \qquad u~\in~\mathbb{R}. \tag{2} $$
In this answer we will only consider OP's integral (1) with the integration contour (2). Here $\epsilon>0$ is a sufficiently small regularization parameter, which the integral (1) does not depend on.

*Next issue is that the square root in the exponential (1) is double-valued. We pick the convention that the imaginary part of a square root $\sqrt{z}$ has the same sign as its argument $z$. In other words, we have a branch cut at the negative ${\rm Re}(z)$-axis. This translates into a branch cut at the imaginary $v$-axis  in the complex $w$-plane for $|v|\geq a$.

*We next deform the integration contour to surround the upper branch cut in the complex $w$-plane. The result is 
$$\begin{align}
I(x)&~~~~=~\int^{v=a}_{v=\infty} \mathrm{d}(iv)~(iv)e^{i\{x(iv)+i\sqrt{v^2-a^2}\}}
+\int_{v=a}^{v=\infty} \mathrm{d}(iv)~(iv)e^{i\{x(iv)-i\sqrt{v^2-a^2}\}} \cr
&~~~~=~-2\int_a^{\infty}\! \mathrm{d}v~ve^{-xv}\sinh\sqrt{v^2-a^2} \cr
&\stackrel{v=t/x+a}{=}~ -\frac{2e^{-xa}}{x}\int_{\mathbb{R}_+}\! \mathrm{d}t~e^{-t}\left(\frac{t}{x}+a\right)\sinh\sqrt{\frac{t}{x}\left(\frac{t}{x}+2a\right)} \cr
&~~~~=~\left\{\begin{array}{rcl}
-\sqrt{2\pi}e^{-xa}\left(\frac{a}{x}\right)^{3/2} \left\{1+O(x^{-1})\right\} &{\rm for}& a~>~0, \cr \cr
-\frac{4}{x^3}\left\{1+O(x^{-1})\right\} &{\rm for}& a~=~0,
,\end{array}\right. \end{align}\tag{3}$$
where we in the last expression gave the leading asymptotic behaviour for $x\to\infty$.
A: Naive reasoning: since you're interested in the case $x \to +\infty$ then the exponential may shrink just into
$$e^{itx}$$
Hence
$$\int_{-\infty}^{+\infty} t\ e^{itx}\ dx = -i\sqrt{2\pi}\delta'(x)$$
Being it the Fourier Transform of the function $t$.
$\delta'$ is the derivative of Dirac Delta function.
A: The integral is equivalently given as the Fourier transform
$$ I_a(x) = \int_{-\infty}^{+\infty} e^{ixt} t e^{-i\sqrt{t^2+a^2}}dt = \mathcal F[\sqrt{2\pi}te^{-i \sqrt{t^2 + a^2}}](-x) $$
If $a=0$ then $e^{-i \sqrt{t^2 + a^2}}$ simplifies to $e^{-i|t|}$ and the integral is analytically solveable:
$$I_0(x) = \frac{-4x}{(x^2-1)^2}$$
Giving $\lim_{x\to\infty} I_0(x) = 0$. I suspect the same holds for all $a$ but I can't prove it quite yet.
