Contour Integral of irrational polynomial from -1 to 1

Question

I've been stuck at htis contour integral problem for a few hours now, and seem to be hitting brick walls.

$$ \int_{-1}^1 \frac{\sqrt{1-x^2}}{1+x^4}dx\,, $$

I tried a trig substitution $x=\cos{\theta}$ but noticed all the poles were the unit circle and I didn't know how to proceed.

$$ \frac{1}{2}\int_{0}^{2\pi} \frac{\sin^2{\theta}}{1+\cos^4{\theta}}d\theta\,, $$

Next I thought maybe a rectangular contour but the contours that go vertically from $1$ to $1+i\infty$ and $-1+i\infty$ to $-1$ didn't seem to cancel, or I wasn't able to show that it does. I manipulated it for awhile.

$$ \int_{0}^\infty \frac{\sqrt{1-(1+iy)^2}}{1+(1+iy)^4}dy + \int_{0}^{-\infty} \frac{\sqrt{1-(1+iy)^2}}{1+(1+iy)^4}dy $$

I naively wrote the trig integral in terms of z, but the denominator is an 8th order polynomial.

$$ i\oint \frac{2z^5-4z^3+2z}{z^8+4z^6+22z^4+4z^2+1} dz $$

I think this might be getting closer to a solution, but how can I find the poles by hand?

Answer 1

Here is a way to compute $$\int_{-\pi/2}^{\pi/2} \frac{\sin^2{\theta}}{1+\cos^4{\theta}}d\theta\,.$$ Let $t=\tan\theta$. Then $$\sin^2\theta=\frac{t^2}{1+t^2},\ \cos^2\theta=\frac{1}{1+t^2},\ d\theta=\frac{dt}{1+t^2}.$$ Plugging them in we get $$\int_{-\infty}^\infty\frac{t^2}{(1+t^2)^2+1}dt.$$ You can find the roots of the denominator easily. So you can use any method you like, say residues or partial fractions, to compute the integral.

Answer 2

As pointed out by @paul garrett, dog-bone contour (dumbbell contour) works perfectly. Indeed, consider the contour $\mathcal{C}$ given as follows:

$\hspace{12em}$

With the principal branch cut, as the radius/band-width of $\mathcal{C}$ goes to zero,

\begin{align*} \oint_{\mathcal{C}} \frac{i\sqrt{z-1}\sqrt{z+1}}{i(z^4+1)} \, dz &\longrightarrow \int_{-1-0^+i}^{1-0^+i} \frac{i\sqrt{z-1}\sqrt{z+1}}{z^4+1} \, dz - \int_{-1+0^+i}^{1+0^+i} \frac{i\sqrt{z-1}\sqrt{z+1}}{z^4+1} \, dz \\ &\quad = 2\int_{-1}^{1} \frac{\sqrt{1-x^2}}{x^4+1} \, dx. \end{align*}

On the other hand, Residue Theorem tells that

\begin{align*} \oint_{\mathcal{C}} \frac{i\sqrt{z-1}\sqrt{z+1}}{z^4+1} \, dz &= - 2\pi i \sum_{a \ : \ a^4 + 1 = 0} \underset{z=a}{\mathrm{Res}} \, \frac{i\sqrt{z-1}\sqrt{z+1}}{z^4+1} \\ &= \pi \sqrt{2(\sqrt{2}-1)}. \end{align*}

(In order to use Residue Theorem, consider a large circle, apply Residue Theorem to the region enclosed by this circle and $\mathcal{C}$, and then let the radius of the circle go to $\infty$.) Therefore

$$ \int_{-1}^{1} \frac{\sqrt{1-x^2}}{x^4+1} \, dx = \frac{\pi}{2} \sqrt{2(\sqrt{2}-1)}. $$