Integral of a sinc squared function over a square root function

[tworitdash]

TL;DR

I want to compute the integral I have posted here ( https://www.physicsforums.com/threads/spectral-domain-double-integral-with-singularities.989141/ ) This is a sub-problem of the actual problem posted there.

I want to find the analytical solution to the integral given below.

$$\int_{-\infty}^{\infty} \frac{ sinc^2(\frac{k_yb}{2})}{\sqrt{k^2 - k_x^2 - k_y^2}}dk_y$$

In other words,

$$\int_{-\infty}^{\infty} \frac{ \sin^2(\frac{k_yb}{2})}{(\frac{k_yb}{2})^2\sqrt{k^2 - k_x^2 - k_y^2}}dk_y$$

Can this be solved analytically?

I have seen that the integral $\int_{-\infty}^{\infty} \frac{ J_0(\frac{k_yb}{2})}{\sqrt{k^2 - k_x^2 - k_y^2}})dk_y$ as $\pi J_0(\frac{b}{4} \sqrt{k^2 - k_x^2}) H_0^{(2)}(\frac{b}{4} \sqrt{k^2 - k_x^2)}$ where $J_0$ is the Bessel function of the first kind of order 0 and $H_0^{(2)}$ is the Hankel function of second kind. Is something similar there for the other integral as well?

 

[anuttarasammyak]

I have a question.
$$\sqrt{k^2-k_x^2-k_y^2}$$
is imaginary number for large k_y. How do you treat it?

 

[anuttarasammyak]

The integral is rewritten as
$$\frac{2}{b} \int_{-\infty}^{+\infty} \frac {1-\cos q}{q^2}\frac{1}{\sqrt{L-(\frac{q}{b})^2}} dq$$
$$=\frac{2}{b} \sum_n a_n \int_{-\infty}^{+\infty} \frac{q^{2n}}{\sqrt{L-(\frac{q}{b})^2}} dq$$
where
$$q=bk_y,L=k^2-k_x^2$$
Each term of Taylor expansion diverges so it does not go well.
Another way might be to divide it as
$$\frac{2}{b} \int_{-\infty}^{+\infty} \frac {1}{q^2}\frac{1}{\sqrt{L-(\frac{q}{b})^2}} dq- \frac{2}{b} \int_{-\infty}^{+\infty} \frac {\cos q}{q^2}\frac{1}{\sqrt{L-(\frac{q}{b})^2}} dq$$
and evaluate each term independently using complex integral for the latter one. But it seems to divert again.

 

[tworitdash]

I have a question.
$$\sqrt{k^2-k_x^2-k_y^2}$$
is imaginary number for large k_y. How do you treat it?

Yes, I use the square root as
$-j \sqrt{-(k^2 - k_x^2 - k_y^2)}$. So, the branch cut appears from INF to -k when real(k_x) is < 0. The other half of the branch cut is from -INF to +k when real(k_x) > 0.

In other words, the square root function makes sure that when $k_y^2 > (k^2 - k_x^2)$, the value of remains purely negative imaginary and when $k_y^2 < (k^2 - k_x^2)$, the value remains purely positive real.

 

[anuttarasammyak]

So,
$$I = 4 \int_{0}^{+\infty} \frac{1-\cos q}{q^2 \sqrt{A^2-q^2}}dq$$
where
$$q=bk_y,\ A^2=b^2(k^2-k_x^2),A>0$$
$$I = 4 \int_{0}^{A} \frac{1-\cos q}{q^2 \sqrt{A^2-q^2}}dq-4i \int_{A}^{+\infty} \frac{1-\cos q}{q^2 \sqrt{q^2-A^2}}dq$$
How about trying substitution
$$q=A \sin \theta$$ for the first integral and
$$q=A \tan \theta$$for the second one ?

 

[anuttarasammyak]

EDIT to post #5

The coefficient of the second integral in RHS is not $- 4i$ but $+4i$ and substitution in it is not $q=A \tan \theta$ but $q=A \sec \theta$.