Evaluate $$I=\int_{-1}^{1} \frac{1}{(1+x^4)\sqrt{1-x^2}}dx$$

by integrating along the following complex contour:

When we take $R\rightarrow 1$ and $ε\rightarrow 0$, the contour integral itself will be $2πi$ times the sum of the 4 residues of the function, which we can get easily enough using L'Hopital's rule, or factorising $1+x^4$ and so on. My real trouble comes when we consider the lower integral along the real line,

$$\int_{1}^{-1} \frac{1}{(1+x^4)\sqrt{1-x^2}}=\int_{1}^{-1} \frac{1}{(1+x^4)\sqrt{1-xe^{2πi}}\sqrt{1+xe^{2πi}}}$$

Where I have included $e^{2πi}$ explicitly since it's the lower path. My intuition is telling me that this piece of the contour shouldn't just be the same as the piece above because of this $e^{2πi}$ inside the square root. Furthermore, if it were the exact same then the two pieces would cancel exactly...

So how we can we separate out the 'effects' of this $e^{2πi}$ to get this piece in the form of constant * $I$?

  • 1
    $\begingroup$ the contour in your pic is incorrect, it should be a dogbone contour. $\endgroup$
    – MathFail
    Jun 13 at 11:13
  • $\begingroup$ @MathFail what do you mean by "incorrect"? Aren't all contours equally valid? (assuming one knows how to use the residue theorem, etc) $\endgroup$
    – Poo2uhaha
    Jun 13 at 11:16
  • $\begingroup$ I mean the shape... $\endgroup$
    – MathFail
    Jun 13 at 11:18
  • $\begingroup$ I don't know where you got $R \to 1$ from, but generally speaking, when working with multi-valued expressions, it's always a good idea to say what intervals your arguments live in (assuming one-to-one correspondence) and how you use those arguments. That way, when you take the limit as $\delta \to 0^+$ (supposing $\delta$ is the small gap length between the branch cut and a line segment), you could see whether the integrals over the two line segments cancel or not. You would also need an indented path around $z=1$. IIRC, maybe using a dogbone contour along with a huge circle around it helps? $\endgroup$ Jun 15 at 8:22

3 Answers 3


To evaluate this integral by means of the residue theorem, we have to create a closed contour, inside which the integrand is a single-valued function. For this purpose we also make necessary cuts. In your case the best option is the usage of a dogbone contour with the cut from $-1$ to $1$, but for the sake of simplicity we can preliminary modify the integral a bit. $$I=\int_{-1}^{1} \frac{1}{(1+t^4)\sqrt{1-t^2}}dt\overset{t^2=x}{=}\int_0^1\frac1{1+x^2}\frac{dx}{\sqrt{x(1-x)}}$$ This substitution will allow us to evaluate only two residues (at $x=\pm i$) instead of four of the initial integral.

We also note that the integral along the dogbone contour (clockwise) with the cut $[0;1]$ gives us $2I$ (the integrals along the arches around $x=0$ and $x=$ tend to zero). Now we consider the integral along the following closed contour $C$:

enter image description here

We added a big circle with the radius $R\to\infty$ and a path $1$, which links the dogbone contour with this circle. Inside the contour the function is single-valued.

Therefore, $$\oint_C\frac1{1+x^2}\frac{dx}{\sqrt{x(1-x)}}=2I+I_1-I_1+I_R=2\pi i\underset{x=e^{\frac{\pi i}2}; \,e^{\frac{3\pi i}2}}{\operatorname{Res}}\frac1{1+x^2}\frac1{\sqrt{x(1-x)}}$$ The integral $I_R\to 0$ as $R\to\infty$, and integrals along the path $1$ cancel each other (we integrate twice, in the opposite directions). $$I=\pi i\underset{x=e^{\frac{\pi i}2}; \,e^{\frac{3\pi i}2}}{\operatorname{Res}}\frac1{(x+i)(x-i)}\frac1{\sqrt{x(1-x)}}=\pi i\left(\frac1{2i}\frac1{\sqrt{e^{\frac{\pi i}2}(1-e^{\frac{\pi i}2})}}-\frac1{2i}\frac1{\sqrt{e^{\frac{3\pi i}2}(1-e^{\frac{3\pi i}2})}}\right)$$ $$=\frac\pi2\left(e^{-\frac{\pi i}4}(1-i)^{-1/2}+e^{\frac{\pi i}4}(1+i)^{-1/2}\right)=\frac\pi{\sqrt2}\Re \,e^{-\frac{\pi i}4}\sqrt{1+i}$$ $$\boxed{\,\,I=\frac\pi22^\frac34\cos\frac\pi8=\frac\pi2\sqrt{1+\sqrt2}\,\,}$$

  • $\begingroup$ Hiya, how come the integrals along the path 1 cancel each other? They have opposite limits (1 to R versus R to 1) but the lower path has $\sqrt{xe^{2πi}} = -\sqrt{x}$ in the denominator of the integrand, so it looks to me like they add instead of cancel. $\endgroup$
    – Poo2uhaha
    Jun 13 at 11:37
  • 1
    $\begingroup$ $\int_{1+r}^R+\int_R^{1+r}=0$, and the cut is only from $0$ to $1$ $\endgroup$
    – Svyatoslav
    Jun 13 at 11:41
  • $\begingroup$ -1, How do you specify the argument angle for each branch cut? Another issue is after doing the $t=x^2$ sub, it changes everything, which doesn't answer the question of OP. $\endgroup$
    – user1026811
    Jun 13 at 12:04
  • $\begingroup$ @GGplay The argument is chosen in a such way that on the upper bank of the cut we get the initial integral. For example, $(1-x)^{-1/2}\to |1-x|^{-1/2}e^{(-\pi i)(-1/2)}=i|1-x|$, when we go (clockwise) around $x=1$ from the upper bunk of the cut to the axis X. I also mentioned that the substitution simplifies evaluation and does not change the ideology (we still use the dogbone contour) - the idea was to show how to use the contour for the evaluation. If you mean that the initial (OP's) contour has to be used - I leave the scene to you. $\endgroup$
    – Svyatoslav
    Jun 13 at 12:31

You have the right idea when you introduce the factor of $\exp(2\pi i)$, but you need to do it in a more subtle way.

When you introduce the $\exp(2\pi i)$ on the lower branch, you should multiply the radicand from each branch by this factor, meaning it's multiplied by $1+x$ in one case and $1-x$ in the other:

$\sqrt{1-x^2}=\sqrt{1+x}\sqrt{1-x}\to\sqrt{(1+x)\exp(2\pi i)}\sqrt{(1-x)\exp(2\pi i)}$

Now this is still not quite right. The argument of the exponential is actually not $2\pi i$. It's $2\pi Ni$, where the winding number $N$ is the number of counterclockwise turns you make when you travel within the countour boundaries from the upper branch to the lower branch (if you go clockwise, the winding number is given a negative sign).

The subtle feature is that those winding numbers could change when you go to different points on the real axis. Let's look at $z=+2$. To get from $+2$ on the upper branch to $+2$ on the lower branch you have to go counterclockwise around both branch points once. So we have at that point on the real axis

$\sqrt{1-2^2}=\sqrt{1+2}\sqrt{1-2}\to\sqrt{(1+2)\exp(2\pi 1i)}\sqrt{(1-2)\exp(2\pi 1i)}=[-\sqrt{1+2}][-\sqrt{1-2}]=\sqrt{1+2}\sqrt{1-2}.$

This says that at $x=2$ the integrand on the lower branch equals the integrand on the upper branch, and since you traverse the branches in opposite directions it cancels itself out.

Now try this at $z=0$. There, the winding number around $-1$ as you go from the upper branch to the lower one is still $1$, but now the winding number around $+1$ is $0$. When I stand at $z=+1$ and view your path from upper-branch $0$ to lower-branch $0$, I see you starting to go around me but then you double back instead of going all the way around. So with the changed winding number around $z=+1$, the lower-branch integrand is changed in sign:

$\sqrt{1-0^2}=\sqrt{1+0}\sqrt{1-0}\to\sqrt{(1+0)\exp(2\pi 1i)}\sqrt{(1-0)\exp(2\pi 0i)}=[-\sqrt{1+0}][\sqrt{1-0}]=-\sqrt{1+0}\sqrt{1-0}.$

With the changed sign, the backwards path along the lower branch no longer cancels the forwards path along the upper branch; instead the contribution is doubled.

So now we see that the contour doubles the integral between $-1$ and $+1$, where the winding numbers are favorable, while canceling it out everywhere else. Therefore the contour integral value you ultimately get doubles the definite integral from $-1$ to $+1$ in real variables, making your answer half the contour value.

The computation is similar to that given by Svyatoslav, and I do not reinvent that wheel here. I merely report the result,



It's not clear to me whether you're obligated to use a specific contour, so I'll suggest another approach. With a substitution of $\displaystyle x=\frac{1-y^2}{1+y^2}$, the integral transforms to

$$\begin{align*} I &= \int_{-1}^1 \frac{dx}{\left(1+x^4\right) \sqrt{1-x^2}} \\ &= \int_0^\infty \frac{\frac{4y}{(1+y^2)^2}\,dy}{\left(1+\frac{(1-y^2)^4}{(1+y^2)^4}\right) \sqrt{1-\frac{(1-y^2)^2}{(1+y^2)^2}}} \\ &= \int_0^\infty \frac{(1+y^2)^3}{1+6y^4+y^8} \, dy \\ &= \frac12 \int_{-\infty}^\infty \frac{(1+y^2)^3}{1+6y^4+y^8} \, dy & . \end{align*}$$

Now integrate over a semicircle in the upper half-plane containing the four poles $\sqrt[4]{3\pm2\sqrt2}\,e^{i\tfrac\pi4}$ and $\sqrt[4]{3\pm2\sqrt2}\,e^{i\tfrac{3\pi}4}$. The residue sum simplifies to $-i \dfrac{\sqrt{1+\sqrt2}}2$ and the result follows.


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