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