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.
Assuming $a\geq 0$,
$$I_2(a) \stackrel{\text{parity}}{=} 2\int_{0}^{+\infty}e^{-\sqrt{t^2+a^2}}\, \mathrm dt \stackrel{t\mapsto a\sinh(z)}{=} 2a\int_{0}^{+\infty}\cosh(z)\,e^{-a\cosh z}\,\mathrm dz\tag{1}$$
leads to:
$$ I_2(a) = 2a\int_{1}^{+\infty}\frac{z}{\sqrt{z^2-1}} e^{-az}\, \mathrm dz = 2a\cdot K_1(a)\tag{2}$$
where $K_1$ is a modified Bessel function of the second kind.
$I_1(a)$ can be computed from Simply Beautiful Art's comment above.
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.
