Integral of a sinc squared function over a square root function

In summary, the conversation discusses finding the analytical solution to the integral \int_{-\infty}^{\infty} \frac{ sinc^2(\frac{k_yb}{2})}{\sqrt{k^2 - k_x^2 - k_y^2}}dk_y , which is rewritten as \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 . It is mentioned that the integral \int_{-\infty}^{\infty} \frac{ J_0(\frac{k
  • #1
tworitdash
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TL;DR 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.
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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  • #2
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
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 divide 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
anuttarasammyak said:
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
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
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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Related to Integral of a sinc squared function over a square root function

1. What is the integral of a sinc squared function over a square root function?

The integral of a sinc squared function over a square root function is a complex mathematical expression that cannot be easily simplified. It involves the integration of a sine function raised to the power of two over a square root function.

2. How is the integral of a sinc squared function over a square root function related to the Fourier transform?

The integral of a sinc squared function over a square root function is closely related to the Fourier transform. In fact, the Fourier transform of a sinc squared function is a square root function, making the integral a key component in the calculation of the inverse Fourier transform.

3. What is the significance of the sinc squared function in signal processing?

The sinc squared function is commonly used in signal processing to represent the frequency response of a low-pass filter. It has a main lobe with a width that is inversely proportional to the cutoff frequency, making it an ideal function for filtering out high frequency components in a signal.

4. Can the integral of a sinc squared function over a square root function be evaluated numerically?

Yes, the integral of a sinc squared function over a square root function can be evaluated numerically using various numerical integration techniques such as the trapezoidal rule or Simpson's rule. However, due to the complexity of the function, the numerical result may not be as accurate as an analytical solution.

5. How is the integral of a sinc squared function over a square root function used in real-world applications?

The integral of a sinc squared function over a square root function has many practical applications in fields such as signal processing, optics, and physics. It is used in the calculation of the frequency response of filters, the design of optical systems, and the analysis of wave propagation phenomena.

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