Finding $\int_{-\infty}^\infty \frac{ \cos(x)}{x^4+1}dx$ using residues

Question

I’m trying to calculate $$\int_{-\infty}^\infty \frac{ \cos(x)}{x^4+1}dx$$ using the residue theorem. I know that the only singularities of the complex function $f(z)=\frac{e^{iz}}{z^4+1}$ which I need to consider are $e^{i\pi /4}$ and $e^{3i\pi /4}$ and that they’re both simple poles, but I’m having trouble calculating the residue of $f(z)$ at each of them (I know how to start but I can’t simplify the expressions I obtain). Is there an easy way to do this?

Answer 1

If $g(z)$ has a single zero at $z_0$, you can use the fact that $$ \lim_{z\to z_0} \frac{z-z_0}{g(z)} = \lim_{z\to z_0} \frac{1}{g'(z)}$$ (from the l'Hopital's rule), so for $z_n \in\{e^{i\pi/4},e^{3i\pi/4}\}$ $$ {\rm Res}\left(\frac{e^{iz}}{z^4+1}, z_n\right) = \lim_{z\to z_n}\left(\frac{e^{iz}(z-z_n)}{z^4+1} \right) = \lim_{z\to z_n}\left(\frac{e^{iz}}{4z^3} \right) = \frac{e^{i z_n}}{4z_n^3}$$

We have $$ \int_{-\infty}^\infty\frac{\cos x}{x^4+1}dx = {\rm Re} \int_{-\infty}^\infty\frac{e^{ix}}{x^4+1}dx = {\rm Re} \left(2\pi i\left(\frac{e^{i z_1}}{4z_1^3} + \frac{e^{i z_2}}{4z_2^3}\right)\right)$$

For $z_1=e^{i\pi/4} = \frac{1+i}{\sqrt{2}}$ we have: \begin{align}{\rm Re}\left(2\pi i\frac{e^{i z_1}}{4z_1^3}\right) & = {\rm Re}\left(\frac{\pi i}{2} e^{\frac{-1+i}{\sqrt{2}}} e^{-3i\pi/4} \right)= \\ &= {\rm Re}\left(\frac{\pi i e^{-\frac{1}{\sqrt{2}}}}{2}\big(\cos\frac{1}{\sqrt{2}} + i\sin\frac{1}{\sqrt{2}} \big)\frac{-1-i}{\sqrt{2}} \right) = \\ &=\frac{\pi e^{-\frac{1}{\sqrt{2}}}}{2\sqrt{2}}\big(\cos\frac{1}{\sqrt{2}} + \sin\frac{1}{\sqrt{2}} \big) \end{align} Similarily, for $z_2=e^{3i\pi/4} = \frac{-1+i}{\sqrt{2}}$ we have: \begin{align}{\rm Re}\left(2\pi i\frac{e^{i z_2}}{4z_2^3}\right) & = {\rm Re}\left(\frac{\pi i}{2} e^{\frac{-1-i}{\sqrt{2}}} e^{-9i\pi/4} \right)= \\ &={\rm Re}\left(\frac{\pi i e^{-\frac{1}{\sqrt{2}}}}{2}\big(\cos\frac{1}{\sqrt{2}} - i\sin\frac{1}{\sqrt{2}} \big)\frac{1-i}{\sqrt{2}} \right) = \\ &=\frac{\pi e^{-\frac{1}{\sqrt{2}}}}{2\sqrt{2}}\big(\cos\frac{1}{\sqrt{2}} + \sin\frac{1}{\sqrt{2}} \big) \end{align} In total, we have $$ \int_{-\infty}^\infty\frac{\cos x}{x^4+1}dx = \frac{\pi e^{-\frac{1}{\sqrt{2}}}}{\sqrt{2}}\big(\cos\frac{1}{\sqrt{2}} + \sin\frac{1}{\sqrt{2}} \big) $$

Answer 2

You need to break up cosine into $(e^{ix}+e^{-ix})/2$ because one of them blows up in the upper half plane and the other blows up in the lower half plane. They need different semicircles that contain different poles.