# Real integral by residues

Does anyone know the answer to this integral? $$\int_{-\infty}^{\infty} \frac{ \cos x dx}{ 1+x^4} \ ?$$

I used $f(z)=\frac{e^{iz}} {1+z^4}$ but calculating the residue at the two singular points of the upper half plane is really complicated. Thanks for your help.

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You might want to use L'Hôpital's rule :) – Nivalth Nov 22 '12 at 11:15
It is not exactly L'Hopitals rule! – Pete Markou Nov 22 '12 at 11:26
Why not, @PeteMarkou? I think it perfectly can be. – DonAntonio Nov 22 '12 at 12:09
@DonAntonio I got it now! – Pete Markou Nov 30 '12 at 10:34

$$z^4=-1=e^{\pi i+2k\pi i}=e^{\pi i(1+2k)}\Longrightarrow z=e^{\frac{\pi i}{4}(1+2k)}\,\,,\,k=0,1,2,3\Longrightarrow$$

the two roots in the upper half plane are

$$z_0=e^{\pi i/4}=\frac{1}{\sqrt 2}(1+i)\;\;,\;\;z_1=e^{3\pi i/4}=\frac{1}{\sqrt 2}(-1+i)$$

So, for example, putting $\,\displaystyle{f(z)=\frac{e^{iz}}{z^4+1}}\,$ , we get

$$Res_{z=z_0}(f)=\lim_{z\to z_0}\frac{(z-z_0)e^{iz}}{z^4+1}\stackrel{\text{L'Hospital}}=\lim_{z\to\frac{1}{\sqrt 2}(1+i)}\frac{e^{iz}}{4z^3}=\frac{e^{\frac{1}{\sqrt 2}(-1+i)}}{2\sqrt 2(-1+i)}$$

Of course, you get the same applying Julián's answer, whose proof uses L'Hospital's rule...

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 No, it does not use L'Hôpital's rule. It uses the definition of derivative, and proves a particular case of L'Hôpital's rule. – Julián Aguirre Nov 25 '12 at 20:25

You may want to use the following result.

Proposition. Let $f$ and $g$ be holomorphic functions on a neighborhood of $a\in\mathbb{C}$ such that $f(a)\ne0$ and $a$ is a simple zero of $g$. Then a is a simple pole of $f/g$ and $$\operatorname{Res}\Bigl(\frac{f}{g},a\Bigr)=\frac{f(a)}{g'(a)}.$$ Proof. $$\operatorname{Res}\Bigl(\frac{f}{g},a\Bigr)=\lim_{z\to a}(z-a)\frac{f(z)}{g(z)}=\lim_{z\to a}\frac{f(z)}{\dfrac{g(z)-g(a)}{z-a}}.$$

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If you don't want to use directly the residues your can also look for the Fourier transform of $\frac{1}{1+x^4}$ and evaluate it at $\omega=1$.

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 Uh? How is this achieved? It looks like an interesting development and I don't know it. What is w, what is the period...? Thanks. – DonAntonio Nov 22 '12 at 12:17 Well, The integral he wants to solve is $F(\omega)=\int_{\infty}^{\infty}\frac{1}{1+x^4}e^{-i\omega x}dx$ at $\omega=1$. $F(\omega)$ is proportional to the Fourier transform of $\frac{1}{1+x^4}$ – Dario Alexander Nov 22 '12 at 12:51 Oh...hehe. Well, yes...I thought something else. Thanks. – DonAntonio Nov 22 '12 at 14:36