Undergrad Changing to Polar coordinates in order to calculate this integral

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The discussion focuses on converting the integral of a function into spherical coordinates, specifically addressing the integration of terms involving cosine and sine functions. The integral includes the expression (cos(kr*cos(theta))+i*sin(kr*cos(theta)))*sin(theta) dtheta, which poses challenges for calculation. The user confirms the validity of the approach with assistance from WolframAlpha, ultimately finding that the imaginary part of the integral is zero while the real part yields a result of 3.67. The conversation highlights the complexities of integrating in polar coordinates and the utility of computational tools in resolving such integrals. Gaussian97 is acknowledged for their support in the process.
Replusz
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I am a bit lost regarding what happens here.
the k^2+m^2 part stays there.
What happens to the exp? and to the d^3k ?

Thank you!
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Try to write 3.66 in spherical coordinates, and perform the integrals of ##\phi## and ##\theta##.
 
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I get a part where I have to integrate (cos(kr*cos(theta))+i*sin(kr*cos(theta)))*sin(theta) dtheta
Which seems terrible
 
Last edited:
OK with the help of wolframalpha I convinced myself that this is indeed what we want.
(imaginary part is 0, real part gives 3.67)

Thank you Gaussian97! :)
 

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