[Q] square of 'sinc function' integral

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The discussion revolves around solving the integral \(\int_{k_o}^0 \frac{\sin^{2}(2x)}{2x^{2}} dx\) to determine the probability density related to momentum eigenstates. The user transformed the integral using the identity \(\sin^{2}(2x) = \frac{1 - \cos(2x)}{2}\) but encountered difficulties due to the singularity at \(x=0\). Attempts to apply integration by parts did not resolve the issue, as the integral diverges. Suggestions include using series expansion for \(\cos(2x)\) to address the singular point. The user seeks guidance on achieving convergence for the integral.
good_phy
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Hi.

I tried to solve some problem that i should get probability density with which eigenstate of

momentum is chosen after momentum measurement by using <\varphi_{k}|\Psi>

I faced some stuck integral problem such as \int_{k_o}^0\frac{sin^{2}(2x)}{2x^{2}}dx

I transformed sin^{2}(2x) = \frac{1 - cos(2x)}{2} so i obtained \int_{k_o}^0\frac{1-cos(2x)}{2x^{2}}dx but i don't know next step because, \int_{k_o}^0\frac{1}{x^2}dx go up to infinity,diverse.

i tried to do partial integral such as \int udv = uv - \int vdu but encountered same problem.

How can i overcome this singular point problem? i convinced that \int_{k_o}^0\frac{sin^{2}(2x)}{2x^{2}}dx

should be solved to convegence because graphic of \frac{sin^{2}(2x)}{2x^{2}}.

Please help me and give me an answer.
 
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i think you can try to write the cos as a series form.
P.S. the third formula in your statements seems...
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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