# Integrating sinc(x)^4 between negative infinity to infinity using complex analysis

 P: 179 You should use the relation $$\sin^4\theta = \frac{3 - 4 \cos 2\theta + \cos 4\theta}{8}$$ It will make your life much easier ^^
 P: 179 What you have written above is absolutely correct. You have 4 paths, let's call them P1,P2,P3,P4. P1 goes from -infinity to -epsilon P2 is a small semicircle from -epsilon to epsilon P3 goes from epsilon to infinity P4 is the large semicirle We already know that the integral over all paths combined is 0. Also integral over P4 is 0. This tells us that $$\int_{P1, P3} f(z) dz = - \int_{P2} f(z) dz$$ Obviously, the combined path P1, P3 is "almost" what we are looking for. Now chose $$f(z) = \frac{1}{8} \frac{3-4e^{2iz}+e^{4iz}}{z^4}$$ The REAL part of $f(z)$ is the function that we want to integrate. Now you need to parametrize P2. choose $$z = \epsilon e^{i \theta}$$ and integrate over $\theta\in[0,\pi]$ while $\epsilon \rightarrow 0$ L'Hôpital might come in handy in your calculations. You might get a complex number as a result. Its real part will be the result you are looking for. Let me know if anything remains unclear.
 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. $$f(z) = \frac{1}{8} \frac{(-8ie^{iθ}e^{2iεe^{iθ}}+4ie^{iθ}e^{4iεe^{iθ}})ie^{iθ}}{4ε^{3}(e^{iθ})^4}$$