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This is a problem that I was trying to solve in preparation for an entrance exam. The first part of the problem was to solve $\int_{-\infty}^\infty \frac{1}{1+x^4} dx$ which is a fairly straight-forward application of complex analysis using a half-circle toy contour. I am not sure however, how I should proceed with this integral. One approach I thought to use was an "expanding rectangle" persae but that didn't seem to work. Any help is appreciated.

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  • $\begingroup$ you can find the indefinite integral by doing $x^4+1 = x^4+2x^2+1-2x^2 = (x^2+1)^2-(\sqrt{2}x)^2 = (x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)$ and then using partial fraction decomposition $\endgroup$ – mathworker21 Jul 25 at 2:10
  • $\begingroup$ but if you wanna use complex analysis, idk, just make a big semicircle along $x$-axis centered at origin with radius tending to $\infty$. $\endgroup$ – mathworker21 Jul 25 at 2:11
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Hint: Consider the integral $$\int_{-\infty}^\infty \frac{e^{ix}}{1+x^4}\,dx.$$ Integrate the corresponding complex function over a closed curve consisting of a line segment $(-R,R)$ and the semicircle from $R$ to $-R$ in the op upper half plane. Since $|e^{iz}|=e^{-y}$ is bounded in the upper half plane, we can conclude that the integral over the semicircle tends to zero. We obtain $$\int_{-\infty}^\infty \frac{e^{ix}}{1+x^4}\,dx=2\pi i \sum_{y>0} \text{Res } \frac{e^{iz}}{1+z^4}$$ Finally, taking the real part gives the answer.

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  • $\begingroup$ Thank you for the explanation. That is very simple in light of the answer to part (a) $\endgroup$ – Adam Martens Jul 25 at 2:37

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