Analytical closed form for a definite integral I am doing a computation in quantum field theory and the following integral occurred to me
$$
   I(a)=\int_{-\infty}^{+\infty}\frac{e^{-a\sqrt{x^2+1}}dx}{x^2+1} \qquad a\ge 0.
$$
I would like to know a closed form for it and the eventual steps to achieve the result.
Thanks.
 A: Let me denote your integral by $I(a)$. Making the change of variables $x=\sinh\varphi$, we get
$$I(a)=\int_{-\infty}^{\infty}\frac{e^{-a\cosh\varphi}d\varphi}{\cosh\varphi}.$$
Next, differentiating this expression with respect to $a$, we get
$$I'(a)=-\int_{-\infty}^{\infty}e^{-a\cosh\varphi}d\varphi=- 2K_0(a),$$
where $K_0(a)$ denotes the Macdonald function. Integrating back, we find that
$$I(a)=I(0)-2\int_0^aK_0(x)dx=\pi-2\int_0^aK_0(x)dx.$$
According to Prudnikov-Brychkov-Marychev (Vol.2, formula 1.12.1.4), the remaining integral is expressed in terms of Macdonald and modified Struve functions, so that
$$I(a)=\pi-2a\left\{K_0(a)+\frac{\pi}{2}\left[K_0(a)\mathbf{L}_1(a)+K_1(a)\mathbf{L}_0(a)\right]\right\}.$$
Presumably this answer does not add anything useful to your integral except that the corresponding special functions have a name, but at least one can be sure that it cannot be simplified further.
A: Here is a closed form in terms of the Meijer $G$-function

$$ I(a)=\int_{-\infty}^{+\infty}\frac{e^{-a\sqrt{x^2+1}}dx}{x^2+1} =\frac{a^2}{4}\,
G^{3, 0}_{1, 3}\left(\frac{{a}^{2}}{4}\, \Bigg\vert\,^{0}_{-\frac{1}{2}, -\frac{1}{2}, -1}\right).$$

A: There is a mistake in the original post: thanks to Jon for pointing it out.
Recognize that the function is even, so you may integrate from 0 to $+\infty$ and multiply by 2, substitute $u=\sqrt{x^2+1}$, and you are left with 
$$
\int_1^{\infty} \frac{e^{-au}}{u^2}\sqrt{u^2-1}du
$$
Which still will not have a closed form involving elementary functions.
Original post:
Recognize that the function is even, so you may integrate from 0 to $+\infty$ and multiply by 2, substitute $u=\sqrt{x^2+1}$, and you are left with 
$$
\int_1^{\infty} \frac{e^{-au}}{u^2}=2E_2(a).
$$
Where $E_2(\cdot)$ is the generalized exponential integral, so a closed form does not exist.
