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**Chris L T521** First note that $\displaystyle \int_0^{\infty}\frac{\,dx}{x^4+1}=\frac{1}{2}\int_ {-\infty}^{\infty}\frac{\,dx}{x^4+1}$

Let $\displaystyle z_1=\frac{\sqrt 2}{2}+i\frac{\sqrt2}{2}$, $\displaystyle z_2=\frac{\sqrt 2}{2}-i\frac{\sqrt2}{2}$, $\displaystyle z_3-\frac{\sqrt 2}{2}+i\frac{\sqrt2}{2}$ and $\displaystyle z_4=-\frac{\sqrt 2}{2}-i\frac{\sqrt2}{2}$

If we select our closed contour to be the semicircle above the real axis, we see that we will have residues at $\displaystyle z_1$ and $\displaystyle z_3$.

So it follows then that

$\displaystyle \text{Res}\!\left(f,z_1\right)=\lim_{z\to z_1}\left(z-z_1\right)\frac{1}{(z-z_1)(z-z_2)(z-z_3)(z-z_4)}$ $\displaystyle =\lim_{z\to z_1}\frac{1}{(z-z_2)(z-z_3)(z-z_4)}=\frac{1}{(i\sqrt{2})(\sqrt{2})(\sqrt{2}(1+i) )}=\frac{i+1}{-4\sqrt{2}}$

$\displaystyle \text{Res}\!\left(f,z_3\right)=\lim_{z\to z_3}\left(z-z_3\right)\frac{1}{(z-z_1)(z-z_2)(z-z_3)(z-z_4)}$ $\displaystyle =\lim_{z\to z_3}\frac{1}{(z-z_1)(z-z_2)(z-z_4)}=\frac{1}{(-\sqrt{2})(\sqrt{2}(i-1))(i\sqrt{2})}=\frac{i-1}{-4\sqrt{2}}$

Thus, by the residue theorem,

$\displaystyle \int_{0}^{\infty}\frac{\,dx}{x^4+1}=\frac{1}{2}\in t_{-\infty}^{\infty}\frac{\,dx}{x^4+1}=\frac{1}{2}\lef t(2\pi i \left(\frac{i+1}{-4\sqrt{2}}+\frac{i-1}{-4\sqrt{2}}\right)\right)$ $\displaystyle =\frac{\pi}{4}\sqrt{2}$

Does this make sense?