How would you integrate this function?

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The discussion focuses on finding the coefficients for a Fourier expansion of the function f(x) = sin(x/2) using the integral (1/π) ∫sin(x/2)sin(nx)dx from -π to π. Participants suggest using integration by parts or the Kronecker delta identity to simplify the process. One user confirms that applying the identity allows for a straightforward solution, while another expresses a preference for using the Kronecker delta for its simplicity. The conversation concludes with a user successfully solving the problem after receiving guidance. This highlights effective methods for calculating Fourier coefficients in mathematical analysis.
Vitani11
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Homework Statement


I need to find the coefficients for a Fourier expansion. Here is the integral I need to solve: (1/π) ∫sin(x/2)sin(nx)dx where the limits are from -π to π. The original function for the expansion is f(x) = sin(x/2)

Homework Equations


None

The Attempt at a Solution


Should I do this in terms of complex numbers? I think that should be my approach but I am not sure. If this is the right approach can you help me get started with setting up the integral?
 
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Vitani11 said:

Homework Statement


I need to find the coefficients for a Fourier expansion. Here is the integral I need to solve: (1/π) ∫sin(x/2)sin(nx)dx where the limits are from -π to π. The original function for the expansion is f(x) = sin(x/2)

Homework Equations


None

The Attempt at a Solution


Should I do this in terms of complex numbers?
I don't think so. The usual approach is integration by parts, twice. Here's a link to a similar example, https://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/trigintsoldirectory/TrigIntSol3.html#SOLUTION 24, problem #24
Vitani11 said:
I think that should be my approach but I am not sure. If this is the right approach can you help me get started with setting up the integral?
 
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use the identity

$$\frac{1}{2\pi} \int_{-\pi} ^{\pi} \sin(nx)\sin(mx) dx = \delta_{nm}$$

and the integral pops right out for you . (i might have the normalization wrong... off the top of my head...)
 
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Woah, so you're saying that using the kronecker delta I can just say it is δn (1/2)? Can you walk me through those steps? Does it have something to do with the dirac delta function/identities? If the kronecker truly makes it simpler then I'd rather use that lol.
 
it is a definition...if you write the sine functions in terms of complex exponential's and do the integrals, it comes out very quickly...
 
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$$\frac{1}{\pi}\int_{-\pi}^{\pi}\sin(nx)\sin(mx)dx = \delta_{nm}$$
 
Yes I've solved this now- thank you
 

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