Question

Prove that dx/x^4 +4=π/8

Answer 1

$\displaystyle\int_0^\infty\frac{\mathrm dx}{x^4+4}$

Consider the complex-valued function

$f(z)=\dfrac1{z^4+4}$

which has simple poles at each of the fourth roots of -4. If $\omega^4=-4$, then

$\omega^4=4e^{i\pi}\implies\omega=\sqrt2e^{i(\pi+2\pi k)/4}$ where $k=0,1,2,3$

Now consider a semicircular contour centered at the origin with radius $R$, where the diameter is affixed to the real axis. Let $C$ denote the perimeter of the contour, with $\gamma_R$ denoting the semicircular part of the contour and $\gamma$ denoting the part of the contour that lies in the real axis.

$\displaystyle\int_Cf(z)\,\mathrm dz=\left\{\int_{\gamma_R}+\int_\gamma\right\}f(z)\,\mathrm dz$

and we'll be considering what happens as $R\to\infty$. Clearly, the latter integral will be correspond exactly to the integral of $\dfrac1{x^4+4}$ over the entire real line. Meanwhile, taking $z=Re^{it}$, we have

$\displaystyle\left|\int_{\gamma_R}\frac{\mathrm dz}{z^4+4}\right|=\left|\int_0^{2\pi}\frac{iRe^{it}}{R^4e^{4it}+4}\,\mathrm dt\right|\le\frac{2\pi R}{R^4+4}$

and as $R\to\infty$, we see that the above integral must approach 0.

Now, by the residue theorem, the value of the contour integral over the entirety of $C$ is given by $2\pi i$ times the sum of the residues at the poles within the region; in this case, there are only two simple poles to consider when $k=0,1$.

$\mathrm{Res}\left(f(z),\sqrt2e^{i\pi/4}\right)=\displaystyle\lim_{z\to\sqrt2e^{i\pi/4}}f(z)(z-\sqrt2e^{i\pi/4})=-\frac1{16}(1+i)$
$\mathrm{Res}\left(f(z),\sqrt2e^{i3\pi/4}\right)=\displaystyle\lim_{z\to\sqrt2e^{i3\pi/4}}f(z)(z-\sqrt2e^{i3\pi/4})=\dfrac1{16}(1-i)$

So we have

$\displaystyle\int_Cf(z)\,\mathrm dz=\int_{\gamma_R}f(z)\,\mathrm dz+\int_\gamma f(z)\,\mathrm dz$
$\displaystyle=0+2\pi i\sum_{z=z_k}\mathrm{Res}(f(z),z_k)$ (where $z_k$ are the poles surrounded by $C$)
$=2\pi i\left(-\dfrac1{16}(1+i)+\dfrac1{16}(1-i)\right)$
$=\dfrac\pi4$

Presumably, we wanted to show that

$\displaystyle\int_0^\infty\frac{\mathrm dx}{x^4+4}=\frac\pi8$

This integrand is even, so

$\displaystyle\int_0^\infty\frac{\mathrm dx}{x^4+4}=\frac12\int_{-\infty}^\infty\frac{\mathrm dx}{x^4+4}=\frac12\frac\pi4=\frac\pi8$

as required.