Integrate Sin(u) over [0, ∞] with Residue Calculus

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SUMMARY

The discussion focuses on integrating the function sin(u) over the interval [0, ∞] using residue calculus, specifically with the integral form ∫_0^∞ sin(u) u^(-B) du, where 0 < B < 2. The presence of the u^B term complicates the integration process, as it prevents the straightforward application of techniques used for simpler integrals like sin(x)/x. Participants highlight the necessity of finding the residue C_-1 and applying the residue theorem, but express challenges in eliminating the imaginary unit i from the calculations.

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kuahji
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\intu^-B sin(u) du, 0<B<2 integrating from 0 to infinity. What is really throwing me off is the condition, I'm not sure why it's there or really what to do with it. Can I just solve this the same way I'd solve sin(x)/x?
 
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Had I not something strange like u^B, I'd solve the problem as follows.
\frac{1}{u^B}(u-\frac{z^3}{6}+...)
I'd then look for C_-1, the residue. Then it's just a simple matter of 2pi(i)*res. But again, having u^B does not allow me to do this. There is a systematic formula I can apply to find the residue, but it does not allow me to eliminate i.
 

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