Definite integral of exponential of square root Does anyone have a clue as to how integrals of the kind
$$
I(a) = \int_{-\infty}^{\infty} dt \frac{e^{-\sqrt{t^2 + a^2}}}{\sqrt{t^2 + a^2}}
$$
or 
$$
I_2(a) = \int_{-\infty}^{\infty} dt e^{-\sqrt{t^2 + a^2}}
$$
can be solved? I realize there is a problem as the square root function has branch cuts, but I do not have a clear idea how to deal with them in this case.
These integrals often arise when calculating matrix elements between electronic wavefunctions in cylindrical coordinates.

For this one $$
I(a) = \int_{-\infty}^{\infty} dt \frac{e^{-\sqrt{t^2 + a^2}}}{\sqrt{t^2 + a^2}}
$$
Note that $(\ln |t+ \sqrt {t^2 + a^2} |)' = \frac {1}{\sqrt {t^2 +a^2}}$
Next notice that $(e^{-\sqrt{t^2 + a^2}})' = -e^{-\sqrt{t^2 + a^2}} * \sqrt{t^2 + a^2}$ 
so $(\ln |t+ \sqrt {t^2 + a^2} |* e^{-\sqrt{t^2 + a^2}})' = \frac {e^{-\sqrt{t^2 + a^2}}}{\sqrt {t^2 +a^2}} + (\ln |t+ \sqrt {t^2 + a^2} |) * -e^{-\sqrt{t^2 + a^2}} * \sqrt{t^2 + a^2}$ hence the $ \int dt \frac{e^{-\sqrt{t^2 + a^2}}}{\sqrt{t^2 + a^2}}
= (\ln |t+ \sqrt {t^2 + a^2} |* e^{-\sqrt{t^2 + a^2}}) -\int (\ln |t+ \sqrt {t^2 + a^2} | * -e^{-\sqrt{t^2 + a^2}} * \sqrt{t^2 + a^2})$ using integration by parts let $dv= (-e^{-\sqrt{t^2 + a^2}} * \sqrt{t^2 + a^2})$ so $v =e^{-\sqrt{t^2 + a^2}} $.
wanted to post this as a comment but it wouldnt fit if its nonsense lemme know and ill delete it.
$ \int dt \frac{e^{-\sqrt{t^2 + a^2}}}{\sqrt{t^2 + a^2}}
= (\ln |t+ \sqrt {t^2 + a^2} |* e^{-\sqrt{t^2 + a^2}}) - (\ln |t+ \sqrt {t^2 + a^2} | * e^{-\sqrt{t^2 + a^2}})  - \int e^{-\sqrt{t^2 + a^2}}$
which jack already solved $\int e^{-\sqrt{t^2 + a^2}}$ above.
