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I'm having difficulties with finding the singularity points:

$$\int_{-\infty}^\infty{\cos(ax)\over x^4 +1}dx, a>0$$

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$x^4=-1$ now use Euler's identity... Drag mouse over grey zone for solutions:

Hence get $\exp(i\frac\pi4), \exp(i\frac{3\pi}4), \exp(-i\frac{\pi}4), \exp(-i\frac{3\pi}4)$

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Use the following contour integral:

$$\oint_C dz \frac{e^{i a z}}{z^4+1}$$

where $C$ is a semicircle in the upper half plane of radius $R$. This integral is equal to

$$\int_{-\infty}^{\infty} dx \frac{e^{i a x}}{x^4+1} + i R \int_0^{\pi} d\theta \, e^{i \theta} \frac{e^{i a R \cos{\theta}} e^{-a R \sin{\theta}}}{R^4 e^{i 4 \theta}+1}$$

As $R\to\infty$, you may show that the second integral vanishes as $\pi/(a R^4)$ by using, e.g., the ML-inequality and the fact that $\sin{\theta} \ge 2 \theta/\pi$ when $\theta \in [0,\pi/2]$.

By the residue theorem, then, the contour integral - and hence the first integral - is equal to $i 2 \pi$ times the sum of the residues of the poles of the integrand inside $C$. These poles are at $z=e^{i \pi/4}$ and $z=e^{i 3 \pi/4}$. The relevant equation is then

$$\int_{-\infty}^{\infty} dx \frac{e^{i a x}}{x^4+1} = i 2 \pi \left [\frac{e^{i a e^{i \pi/4}}}{4 e^{i 3 \pi/4}} + \frac{e^{i a e^{i 3 \pi/4}}}{4 e^{i 9 \pi/4}} \right ] = i \frac{\pi}{2} e^{-a/\sqrt{2}} \left (e^{-i 3 \pi/4} e^{i a/\sqrt{2}} + e^{-i \pi/4} e^{-i a/\sqrt{2}} \right )$$

The quantity in parentheses is

$$-\frac1{\sqrt{2}} \left [(1+i) \left (\cos{\frac{a}{\sqrt{2}}}+i \sin{\frac{a}{\sqrt{2}}}\right ) - (1-i) \left (\cos{\frac{a}{\sqrt{2}}}-i \sin{\frac{a}{\sqrt{2}}}\right ) \right ] $$

which simplifies to

$$-i \sqrt{2} \left (\cos{\frac{a}{\sqrt{2}}} + \sin{\frac{a}{\sqrt{2}}}\right ) $$

Thus, exploiting the symmetry in the integral, we finally have

$$\int_{-\infty}^{\infty} dx \frac{\cos{a x}}{x^4+1} = \frac{\pi}{\sqrt{2}} e^{-a/\sqrt{2}}\left (\cos{\frac{a}{\sqrt{2}}} + \sin{\frac{a}{\sqrt{2}}}\right ) $$

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  • $\begingroup$ What you did for the previous one inspired me a lot. I remember how much I have been impressed at the time. Unfortunately, because of my condition, I was not able to type an as nice answer as your. Thanks and cheers. $\endgroup$ – Claude Leibovici Jan 30 '14 at 8:29
  • $\begingroup$ @ClaudeLeibovici: Many thanks, I am grateful that you have been able to work out this example on your own from an answer I provided. That is why I spend so much time writing up these solutions. $\endgroup$ – Ron Gordon Jan 30 '14 at 8:39
  • $\begingroup$ May I say that, thanks to MSE, I learnt a lot from you (an some others) ? Thanks. $\endgroup$ – Claude Leibovici Jan 30 '14 at 8:40
  • $\begingroup$ I'm sure you're completely aware of this, but if you integrate the same function around a quadrant, you only have to deal with one residue and you can extract both $\int_{0}^{\infty} \frac{\cos ax}{1+x^{4}} \ dx$ and $\int_{0}^{\infty} \frac{\sin ax - e^{-ax}}{1+x^{4}} \ dx$. $\endgroup$ – Random Variable Jan 30 '14 at 15:35
  • $\begingroup$ @RandomVariable: well aware of it, but thanks for pointing out. Sometimes you have to decide between the cleverness of the integrand/contour and using only one pole, or just using the obvious integrand/contour and using two poles. I chose the latter here. $\endgroup$ – Ron Gordon Jan 30 '14 at 15:46
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Doing something similar to what Ron Gordon gave as an answer to the post mentioned by Git Gud, I arrived to a rather simple expression which write
$$\frac{\pi e^{-\frac{|a|}{\sqrt{2}}} \left(\sin \left(\frac{|a|}{\sqrt{2}}\right)+\cos \left(\frac{a}{\sqrt{2}}\right)\right)}{\sqrt{2}}$$
which is valid if $a\in \mathbb{R}$.

A side result I found amazing : the integral of the above axpression between $0$ and $\infty$ is equal to $\pi$.

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