# Showing that parts of a contour integral vanish

I'm trying to compute the integral $\int_{0}^{\infty}\frac{x}{x^6+1}dx$ using contours in the complex plane.

To do this I take the function $f(z)=\frac{z}{z^6+1}$ and integrate along the contour consisting of the positive real line, a circle of infinite radius going counter clockwise from the positive real axis to the positive imaginary axis, and a line going back to the origin. Since there is a singularity on the imaginary axis I introduce an additional contour going clockwise around it. Using residues I can compute the integral of $f(z)$ along this contour, however, in order to equate it to the real integral I must show that the integrals along the imaginary axis vanish.

Can this be done by simply applying the ML estimate, as I do to show that the integral along the semicircle vanishes, or am I missing something?

There is a pole at $z=i$ and so we cannot close the contour along the imaginary axis.

Rather, let's chose a contour $C_R$ to be comprised on $(1)$ the real line segment from $z=0$ to $z=R$, $(2)$ the circular arc described parametrically by $z=Re^{i\phi}$, $\phi \in [0, \pi/3]$, and the straight-line segment from $z=Re^{i\pi/3}$ to $z=0$.

If $R>1$, then only the pole at $z=e^{i\pi/6}$ is enclosed by $C_R$. We thus have from the residue theorem

\begin{align} \oint_{C_R} \frac{z}{z^6+1}\,dz&=2\pi i \text{Res}\left(\frac{z}{z^6+1}, z=e^{i\pi/6}\right)\\\\ &=2\pi i \frac{e^{-2\pi/3}}{6} \tag1 \end{align}

We also have

\begin{align} \oint_{C_R} \frac{z}{z^6+1}\,dz&=\int_0^R \frac{x}{x^6+1}\,dx+\int_0^{\pi/3}\frac{Re^{i\phi}}{R^6e^{i6\phi}+1}\,iRe^{i\phi}\,d\phi+\int_R^0 \frac{xe^{i2\pi/3}}{x^6+1}\,dx\\\\ &=\left(1-e^{i2\pi/3}\right)\int_0^R \frac{x}{x^6+1}\,dx++\int_0^{\pi/3}\frac{Re^{i\phi}}{R^6e^{i6\phi}+1}\,iRe^{i\phi}\,d\phi\\\\ &=-i2e^{i\pi/3}\sin(\pi/3)\int_0^R \frac{x}{x^6+1}\,dx++\int_0^{\pi/3}\frac{Re^{i\phi}}{R^6e^{i6\phi}+1}\,iRe^{i\phi}\,d\phi\tag2 \end{align}

The second integral on the right-hand side of $(2)$ approaches $0$ as $R\to \infty$ and the first integral approaches the integral of interest.

Equating $(1)$ and $(2)$ after letting $R\to \infty$ yields

$$\int_0^\infty \frac{x}{x^6+1}\,dx=\frac{\pi}{3\sqrt 3}$$

• This does indeed give the right result. The trick seems to be in choosing the final contour such that its parametrization gives some factor of the integral we want to compute. – fizicar Nov 7 '17 at 0:28
• @fizicar Indeed. – Mark Viola Nov 7 '17 at 0:56