$$\int_{0}^{\infty} x^{\nu} \frac{e^{-\sqrt{x^2+a^2}}}{\sqrt{x^2+a^2}} \, dx$$
Is it possible to calculate this for $a>0$ and $\nu=0, 2$ ?
I think the result seems to include exponential integral function, but I failed to find the answer from the integration table.
I would be very grateful if you could share some of the good integration skills, ideas, or any advice.
Let $x=a \sinh t$, then $$I_1=\int_{0}^{\infty} \frac{e^{-\sqrt{x^2+a^2}}}{\sqrt{x^2+a^2}} dx= \int_{0}^{\infty} e^{-a \cosh t} dt=K_0(a),$$ where $K_{\nu}(x)$ is cylindrical modified Bessel function of order $\nu$. Next, $$I_1=\int_{0}^{\infty} x^2 \frac{e^{-\sqrt{x^2+a^2}}}{\sqrt{x^2+a^2}} dx=\frac{a^2}{2}\int_{0}^{\infty} (\cosh 2t-1)~ e^{-a\cosh t} dt=\frac{a^2}{2} [K_2(a)-K_0(a)]$$ See for $K_{\nu}(z):$
https://en.wikipedia.org/wiki/Bessel_function