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 P: 13 Integrating sinc(x)^4 between negative infinity to infinity using complex analysis Thank you for the reply. I will try integrating with the relation you've proposed. As for the other method, the one I started with, the limit tends to infinity in my calculations. If z=εe^(iθ), then dz=iεe^(iθ) and the equation you've written becomes $$f(z) = \frac{1}{8} \frac{(3-4e^{2iεe^{iθ}}+e^{4iεe^{iθ}})iεe^{iθ}}{(εe^{iθ})^4}$$ You can factorize your epsilon above to have ε^3 in the bottom. Now you have a form of 0/0, which is alright because you can use L'Hôpital's. After that however, we have something that tends to infinity as ε-->0. $$f(z) = \frac{1}{8} \frac{(-8ie^{iθ}e^{2iεe^{iθ}}+4ie^{iθ}e^{4iεe^{iθ}})ie^{iθ}}{4ε^{3}(e^{iθ})^4}$$