[tex]\displaystyle\int_0^\infty\frac{\mathrm dx}{x^4+4}[/tex]
Consider the complex-valued function
[tex]f(z)=\dfrac1{z^4+4}[/tex]
which has simple poles at each of the fourth roots of -4. If [tex]\omega^4=-4[/tex], then
[tex]\omega^4=4e^{i\pi}\implies\omega=\sqrt2e^{i(\pi+2\pi k)/4}[/tex] where [tex]k=0,1,2,3[/tex]
Now consider a semicircular contour centered at the origin with radius [tex]R[/tex], where the diameter is affixed to the real axis. Let [tex]C[/tex] denote the perimeter of the contour, with [tex]\gamma_R[/tex] denoting the semicircular part of the contour and [tex]\gamma[/tex] denoting the part of the contour that lies in the real axis.
[tex]\displaystyle\int_Cf(z)\,\mathrm dz=\left\{\int_{\gamma_R}+\int_\gamma\right\}f(z)\,\mathrm dz[/tex]
and we'll be considering what happens as [tex]R\to\infty[/tex]. Clearly, the latter integral will be correspond exactly to the integral of [tex]\dfrac1{x^4+4}[/tex] over the entire real line. Meanwhile, taking [tex]z=Re^{it}[/tex], we have
[tex]\displaystyle\left|\int_{\gamma_R}\frac{\mathrm dz}{z^4+4}\right|=\left|\int_0^{2\pi}\frac{iRe^{it}}{R^4e^{4it}+4}\,\mathrm dt\right|\le\frac{2\pi R}{R^4+4}[/tex]
and as [tex]R\to\infty[/tex], we see that the above integral must approach 0.
Now, by the residue theorem, the value of the contour integral over the entirety of [tex]C[/tex] is given by [tex]2\pi i[/tex] times the sum of the residues at the poles within the region; in this case, there are only two simple poles to consider when [tex]k=0,1[/tex].
[tex]\mathrm{Res}\left(f(z),\sqrt2e^{i\pi/4}\right)=\displaystyle\lim_{z\to\sqrt2e^{i\pi/4}}f(z)(z-\sqrt2e^{i\pi/4})=-\frac1{16}(1+i)[/tex]
[tex]\mathrm{Res}\left(f(z),\sqrt2e^{i3\pi/4}\right)=\displaystyle\lim_{z\to\sqrt2e^{i3\pi/4}}f(z)(z-\sqrt2e^{i3\pi/4})=\dfrac1{16}(1-i)[/tex]
So we have
[tex]\displaystyle\int_Cf(z)\,\mathrm dz=\int_{\gamma_R}f(z)\,\mathrm dz+\int_\gamma f(z)\,\mathrm dz[/tex]
[tex]\displaystyle=0+2\pi i\sum_{z=z_k}\mathrm{Res}(f(z),z_k)[/tex] (where [tex]z_k[/tex] are the poles surrounded by [tex]C[/tex])
[tex]=2\pi i\left(-\dfrac1{16}(1+i)+\dfrac1{16}(1-i)\right)[/tex]
[tex]=\dfrac\pi4[/tex]
Presumably, we wanted to show that
[tex]\displaystyle\int_0^\infty\frac{\mathrm dx}{x^4+4}=\frac\pi8[/tex]
This integrand is even, so
[tex]\displaystyle\int_0^\infty\frac{\mathrm dx}{x^4+4}=\frac12\int_{-\infty}^\infty\frac{\mathrm dx}{x^4+4}=\frac12\frac\pi4=\frac\pi8[/tex]
as required.
