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How to get the integral result?

  • #1
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6

Homework Statement


I am studying Gerry's <Introductory Quantum Optics>, in which there is an integral (Eq. 4.37)
$$\intop_{-infinity}^{+infinity}\frac{[sin(\triangle t/2)]^{2}}{\triangle^{2}}d\triangle=\frac{\pi}{2}t.$$
I don't know how to get the result of the right side.

Homework Equations


I have no idea.

The Attempt at a Solution


I tried like the following
$$\intop_{-infinity}^{+infinity}\frac{[sin(\triangle t/2)]^{2}}{\triangle^{2}}d\triangle=\intop_{-infinity}^{+infinity}\frac{1-cos(\triangle t)}{2\triangle^{2}}d\triangle$$
But don't know what to do next.
 

Answers and Replies

  • #2
BvU
Science Advisor
Homework Helper
2019 Award
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Matter of googling "integral sin^2x/x^2" to see our math colleagues have a thread on this one....

But for more specific questions PF is of course available with a :smile:
 
  • #3
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
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Homework Statement


I am studying Gerry's <Introductory Quantum Optics>, in which there is an integral (Eq. 4.37)
$$\intop_{-infinity}^{+infinity}\frac{[sin(\triangle t/2)]^{2}}{\triangle^{2}}d\triangle=\frac{\pi}{2}t.$$
I don't know how to get the result of the right side.

Homework Equations


I have no idea.

The Attempt at a Solution


I tried like the following
$$\intop_{-infinity}^{+infinity}\frac{[sin(\triangle t/2)]^{2}}{\triangle^{2}}d\triangle=\intop_{-infinity}^{+infinity}\frac{1-cos(\triangle t)}{2\triangle^{2}}d\triangle$$
But don't know what to do next.
In questions of this type you should always try to simplify first, before doing anything else. So if we use the notation ##w## instead of your ##\Delta## and ##a## instead of your ##t/2##, we have an integral of the form
$$I(a) = \int_{-\infty}^{\infty} \frac{\sin^2(a w)}{w^2} \, dw.$$
Change variables to ##u = a w## to get ##I(a) = a K##, where
$$K= \int_{-\infty}^{\infty} \frac{\sin^2 u}{u^2} \, du.$$
Now ##K## is just some constant; it happens to equal ##\pi##, but that is another issue.
 
  • #4
hunt_mat
Homework Helper
1,724
16
I think there is a simple solution. Take:

[tex]
I(a)=\int_{-\infty}^{\infty}\frac{\sin^{2}ax}{x^{2}}dx
[/tex]

Now differentiate w.r.t. [itex]a[/itex] to find that:

[tex]
I'(a)=2\int_{-\infty}^{\infty}\frac{\sin 2ax}{2x}dx
[/tex]

Now the integral on the RHS can be calculated using residue calculus. It's result will be a function of [itex]a[/itex], the result is found out via integration.
 

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