Asymptotic Behavior of Modified Bessel Functions

[stevendaryl]

Does anyone know whether the following integral has a closed-form solution? If not, is anything known about the asymptotic behavior?

$f(x) = \int_{-\infty}^{+\infty} \frac{e^{iux}}{\sqrt{u^2 + 1}} du$

 

[Paul Colby]

Does it help that $f''(x)-f(x)=\delta(x)$?

Oops missed the square root

 

[Charles Link]

I don't know the complex variable residue theory as well as I should, but for x>0, I think the contour closes in the upper half plane and has a single pole at z=+i inside the contour so that $f(x)=2 \pi i exp(-x)$. For x<0, the contour would close in the lower half plane with a pole at z=-i and the counterclockwise path would give a minus sign, so that $f(x)=-2 \pi i exp(+x)$. I'm not sure about this function with the sqrt. in the denominator is considered to have simple poles. Perhaps @micromass can answer that. editing... and I missed the term of sqrt (2i) in the denominator for the part without the pole. Will need to research this further...editing ... multiplying numerator and denominator by sqrt (u^2+1) gives the function simple poles, but I'm still at the drawing board=perhaps there is a simple solution... editing... thought I had a possible answer, but still at the drawing board...

 

[Hawkeye18]

It is modified Bessel functions of the second kind $K_0(x)$ (I assume $z=x$ in your integral), see, for example, here, equation (6). You should find a lot of information about Bessel functions (as well as about all other special functions) in Abramowitz and Stegun.

 

[stevendaryl]

It is modified Bessel functions of the second kind $K_0(x)$ (I assume $z=x$ in your integral), see, for example, here, equation (6). You should find a lot of information about Bessel functions (as well as about all other special functions) in Abramowitz and Stegun.

Thanks. I should have noticed that the imaginary part of my integral vanishes by symmetry.