Integral of a sinc squared function over a square root function

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  • #1
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Summary:

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.

Main Question or Discussion Point

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

[tex] \int_{-\infty}^{\infty} \frac{ sinc^2(\frac{k_yb}{2})}{\sqrt{k^2 - k_x^2 - k_y^2}}dk_y [/tex]

In other words,

[tex] \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 [/tex]

Can this be solved analytically?

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

  • #2
anuttarasammyak
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I have a question.
[tex]\sqrt{k^2-k_x^2-k_y^2}[/tex]
is imaginary number for large k_y. How do you treat it?
 
  • #3
anuttarasammyak
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The integral is rewritten as
[tex]\frac{2}{b} \int_{-\infty}^{+\infty} \frac {1-\cos q}{q^2}\frac{1}{\sqrt{L-(\frac{q}{b})^2}} dq[/tex]
[tex]=\frac{2}{b} \sum_n a_n \int_{-\infty}^{+\infty} \frac{q^{2n}}{\sqrt{L-(\frac{q}{b})^2}} dq[/tex]
where
[tex]q=bk_y,L=k^2-k_x^2[/tex]
Each term of Taylor expansion diverges so it does not go well.
Another way might be to devide it as
[tex]\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[/tex]
and evaluate each term independently using complex integral for the latter one. But it seems to divert again.
 
  • #4
41
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I have a question.
[tex]\sqrt{k^2-k_x^2-k_y^2}[/tex]
is imaginary number for large k_y. How do you treat it?
Yes, I use the square root as
[itex] -j \sqrt{-(k^2 - k_x^2 - k_y^2)}[/itex]. 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 [itex] k_y^2 > (k^2 - k_x^2) [/itex], the value of remains purely negative imaginary and when [itex] k_y^2 < (k^2 - k_x^2)[/itex], the value remains purely positive real.
 
  • #5
anuttarasammyak
Gold Member
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So,
[tex]I = 4 \int_{0}^{+\infty} \frac{1-\cos q}{q^2 \sqrt{A^2-q^2}}dq[/tex]
where
[tex]q=bk_y,\ A^2=b^2(k^2-k_x^2),A>0[/tex]
[tex]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[/tex]
How about trying substitution
[tex]q=A \sin \theta[/tex] for the first integral and
[tex]q=A \tan \theta[/tex]for the second one ?
 
  • #6
anuttarasammyak
Gold Member
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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##.
 
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