Compute $\int_0^\infty \frac{dx}{x^5+1}$ using a contour in the upper half complex plane that encloses one of the roots of $z^5+1=0$

Question

Question:

Compute $\int_0^\infty \frac{dx}{x^5+1}$ using a contour in the upper half complex plane that encloses one of the roots of $z^5+1=0$.

Hint: The contour should consist of the interval $[1,R]$, a circle of radius $R$, and part of a ray connecting a point on the circle with the origin.

I've done a few questions like this before where the resulting segments of the contour all go to zero in the limit except for the portion on the real line - so I understand the concept. However, the hints for the contour above confuse me. Shouldn't we require a 4th segment that connects the origin to the point $1$ on the positive real line?

Answer 1

I find the hint somewhat confusing. The limits of the integral suggest part of the contour should be the segment $[0, R]$, $R > 1$, not $[1, R]$.

Now, we shouldn't necessarily expect the integral over each of the other constituent arcs in the contour to go to zero, as any such arc has one end at $0$, where the integrand is positive. We can avoid the problem this might otherwise cause by observing that the integrand is invariant under rotation about the origin by a multiple of $\frac{2\pi}{5}$, which in particular suggests we choose another arc of the contour to be the image of $[0, R]$ under, e.g., anticlockwise rotation by $\frac{2 \pi}{5}$, namely, the line segment connecting $0$ and $R e^{2 \pi i / 5}$.

Then, it's particularly easy to estimate the value of the integral over the circular arc of radius $R$ centered at the origin and connecting $R$ and $R e^{2 \pi i / 5}$, and these three (appropriately oriented) arcs together form a closed contour.