# Contour Integral of irrational polynomial from -1 to 1

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?

• Did you try the "dumb-bell" contour? (small circle around +1, then over to -1, small circle around it, back to +1). Since your integrand is holomorphic outside this, you can move the dumb-bell contour out to infinity, seeing that it is 0. At the same time, the values of it on $(-1,+1)$ differ... Similar trick to keyhole/Hankel contour? – paul garrett Sep 7 '18 at 21:37

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

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

• Thank you! I have one question though, why don't your integrals annihilate each other? I noticed you introduced an $\frac{i}{i}$ in the first integral? – walczyk Sep 8 '18 at 1:11
• @walczyk, That's the beauty of branch cut. Test main observation is that $\sqrt{-x\pm0^+i}=\pm i\sqrt{x}$ for $x\geq0$ when using the principal branch cut. – Sangchul Lee Sep 8 '18 at 1:23
• @SanchulLee Ahh I remember now, it picks up a $e^{i\pi}$ around the cut no? – walczyk Sep 8 '18 at 1:28
• Tes. That's true :) – Sangchul Lee Sep 8 '18 at 1:34
• I was just thinking... this contour, doesn't it not encircle any poles?? Or, rather, you are saying it's possible to deform it into a giant circle? So it encloses every pole? – walczyk Sep 8 '18 at 4:34

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.

• is there a name for this substitution? It's extremely clever. – walczyk Sep 12 '18 at 16:14
• I don't know. The substitution $t=\tan\frac{\theta}{2}$ is more famous which solves all integrals of rational functions of trigonometric functions. But in many cases $t=\tan\theta$, $t=\sin\theta$ or $t=\cos\theta$ will suffice. – Eclipse Sun Sep 12 '18 at 17:34
• how can we show that the integral remains unchanged when we change the integrand's limits from [0,pi] to [-pi/2,pi/2]? I think there must be a simple way of showing it, but I dunno how exactly – walczyk Sep 12 '18 at 21:37
• You can see that the integrand has period $\pi$. So you can integrate over any interval of length $\pi$. – Eclipse Sun Sep 13 '18 at 0:01