analysiscomplex-analysiscontour-integrationdefinite integralsintegration
Q5 Use contour integration to evaluate $$\int_{-1}^{1}\frac{\sqrt[]{1-x^2}}{x^2 + 1}dx$$
I've been trying to use a similar method to the one given in https://en.wikipedia.org/wiki/Contour_integration#Example_6_–_logarithms_and_the_residue_at_infinity but with a branch cut on [-1,1]. However I don't really understand what to do.
Consider $f(\omega) = \omega^{-3/4} (x-\omega)^{-3/4} (1-\omega)^{-3/4}$ and $ I(x) = \oint_{\vert \omega \vert = x} f(\omega) \mathrm{d} \omega$.
The function $f(\omega)$ is discontinuous at $\omega = -x$ along $\vert \omega \vert = x$ and has integrable singularity at $\omega = x$. Making a change of variables, $\omega = x z$: $$ I(x) = \frac{1}{\sqrt{x}} \oint_{\vert z \vert = 1} z^{-3/4} (1-z)^{-3/4} (1-x z)^{-3/4} \mathrm{d} z = \frac{1}{\sqrt{x}} \oint_{\vert z \vert = 1} h\left(z\right) \, \mathrm{d} z $$
Let $\mathcal{C}$ denote the circle $\vert z \vert = 1$. Let $C_{-1, \delta}$ denote segment of $\mathcal{C}$ which crosses the negative axis and of length $2 \pi \delta$, with $\pi \delta$ above and $\pi \delta$ below the negative axis.
It is clear that integral along $\mathcal{C}_{-1,\delta}$ is vanishing as $\delta \to 0$: $$ \begin{eqnarray} \left\vert \int_{\mathcal{C}_{-1,\delta}} z^{-3/4} (1-z)^{-3/4} (1-x z)^{-3/4} \mathrm{d} z \right\vert &\le& (2(1+x))^{-3/4} \left\vert \int_{\mathcal{C}_{-1,\delta}} z^{-3/4} \mathrm{d} z \right\vert \\ &=& \left(2 (1+x) \right)^{-3/4} \frac{8\sqrt{2}}{7} \sin\left(\frac{7 \delta}{8}\right) \left( \cos\left(\frac{7 \delta}{8}\right) - \sin\left(\frac{7 \delta}{8}\right) \right) \\ &\le& \sqrt{2} \delta \left(2 (1+x) \right)^{-3/4} \end{eqnarray} $$
Let's complete $\mathcal{C} \backslash \mathcal{C}_{-1,\delta}$ with integration along $(-1,0)$ above the axis and then along $(0,-1)$ below the axis so as to complete the contour, and call the completed contour $\mathcal{L}$. Then $$ \begin{eqnarray} \oint_{\mathcal{C}} h\left(z\right) \, \mathrm{d} z &=& \oint_{\mathcal{L}} h\left(z\right) \, \mathrm{d} z + \int_0^1 \left( h\left(-y - i \epsilon \right) - h\left(-y + i \epsilon \right) \right)\, \mathrm{d} y \\ &=& \oint_{\mathcal{L}} h\left(z\right) \, \mathrm{d} z + \left( \mathrm{e}^{i \frac{3 \pi}{4}} - \mathrm{e}^{-i \frac{3 \pi}{4}} \right) \int_0^1 y^{-3/4}(1+y)^{-3/4}(1+x y)^{-3/4} \mathrm{d} y \end{eqnarray} $$ The claim is that the principal value of $\oint_\mathcal{L} h(z) \mathrm{d} z = 0$, so we get $$ I(x) = \frac{2 i \sin\left( 3/4 \pi \right)}{\sqrt{x}} \int_0^1 y^{-3/4}(1+y)^{-3/4}(1+x y)^{-3/4} \mathrm{d} y $$
Now, let's check this with quadratures:
Notice that the purported answer you gave in your post can not be correct, as it is not purely imaginary.
The main question is how do we want to define $\frac1{\sqrt{x^2-1}}$ along $[-1,1]$? There are two possibilities: $\frac1{\sqrt{x^2-1}}=\frac{\pm i}{\sqrt{1-x^2}}$. Once we decide that, things are pretty simple.
We can define $$ \log\left(\frac{z+1}{z-1}\right) =\log(3)+\int_2^z\left(\frac1{w+1}-\frac1{w-1}\right)\mathrm{d}w\tag{1} $$ where the integral is evaluated along any path which does not intersect $[-1,1]$. A closed path avoiding $[-1,1]$ will circle both poles an equal number of times and the residues will cancel.
Therefore, $(1)$ defines $\log\left(\frac{z+1}{z-1}\right)$ with a branch cut along $[-1,1]$. Using $(1)$, we can define $$ \frac1{\sqrt{z^2-1}}=\frac1{z+1}e^{\frac12\log\left(\frac{z+1}{z-1}\right)}\tag{2} $$ According $(2)$, the integrand along the top of $[-1,1]$ is $\frac{-i}{\sqrt{1-z^2}}$ and along the bottom of $[-1,1]$ is $\frac{i}{\sqrt{1-z^2}}$. The integral around the two dumbbell ends vanish as their size gets smaller. Thus, the integral counter-clockwise along the whole dumbbell is $$ 4i\int_0^1\frac{\mathrm{d}x}{\sqrt{1-x^2}}\tag{3} $$ The integral of $\frac1{\sqrt{z^2-1} }$, as defined in $(2)$, counter-clockwise around a circle of essentially infinite radius is $2\pi i$.
Cauchy's Integral Theorem says that the integral around the dumbbell and a circle of essentially infinite radius are the same. Thus, $(3)$ says $$ 4i\int_0^1\frac{\mathrm{d}x}{\sqrt{1-x^2}}=2\pi i\tag{4} $$ We are back to the question I raised at the beginning: how to define $\frac1{\sqrt{x^2-1}}$ along $[-1,1]$.
If we look at $\frac1{\sqrt{x^2-1}}$ as $\frac1{\sqrt{z^2-1}}$ along the top of $[-1,1]$, then $\frac1{\sqrt{x^2-1}}=\frac{-i}{\sqrt{1-x^2}}$ and we get $$ \int_0^1\frac{\mathrm{d}x}{\sqrt{x^2-1}}=-i\frac\pi2\tag{5} $$ If we look at $\frac1{\sqrt{x^2-1}}$ as $\frac1{\sqrt{z^2-1}}$ along the bottom of $[-1,1]$, then $\frac1{\sqrt{x^2-1}}=\frac{i}{\sqrt{1-x^2}}$ and we get $$ \int_0^1\frac{\mathrm{d}x}{\sqrt{x^2-1}}=i\frac\pi2\tag{6} $$
Best Answer
The integral around the classical "dogbone" contour implicates the residues from the poles at $z=i$ and $z=-i$ along with the residue at $\infty$.
First, we cut the plane with branch cuts that begin at $z=\pm 1$ and extend to the point at $\infty$ along the negative real axis. We write $\sqrt{1-z^2}=-i\sqrt{z+1}\sqrt{z-1}$ and select the branches of the square roots such that if we approach the line segment $[-1,1]$ from the upper half plane, then $\arg(\sqrt{1-z^2})=0$. Proceeding, we select
$$-\pi<\arg(z+1)\le \pi \tag 1$$
and
$$-\pi<\arg(z-1)\le \pi \tag 2$$
Note that given $(1)$ and $(2)$, we see that as we approach $[-1,1]$ from the lower half plane, then $\arg(\sqrt{1-z^2})=-\pi$.
Let $C$ be the classical dogbone contour comprised of the $(i)$ line segment from $-1+$ to $1$ approaching from the upper half plane, $(ii)$ line segment from $1+$ to $-1$ approaching from the lower half plane, $(iii)$ a circle of radius $\epsilon$ and center $-1$, and $(iv)$ a circle of radius $\epsilon$ and center $1$. In the limit as $\epsilon\to 0$, the contributions to the value of the integral from integrating over $(iii)$ and $(iv)$ vanish. Hence, we have
$$\begin{align} \lim_{\epsilon\to 0^+}\oint_C \frac{\sqrt{1-z^2}}{z^2+1}\,dx&=\int_{-1}^1 \frac{\sqrt{1-x^2}}{1+x^2}\,dx+\int_1^{-1}\frac{-\sqrt{1-x^2}}{1+x^2}\,dx\\\\ &=2\int_{-1}^1 \frac{\sqrt{1-x^2}}{1+x^2}\,dx\tag 3 \end{align}$$
In addition, we have
$$\begin{align}\lim_{\epsilon\to 0^+}\oint_C \frac{\sqrt{1-z^2}}{z^2+1}\,dx&=2\pi i\text{Res}\left(\frac{\sqrt{1-z^2}}{1+z^2}, z=\pm i,\infty\right) \\\\ &=2\pi i\left(\frac{\sqrt 2}{2i}+\frac{-\sqrt 2}{-2i}+i\right)\\\\ &=\pi (2\sqrt 2-2)\tag 4 \end{align}$$
Equating $(3)$ and $(4)$ and solving for the integral of interest yields
$$\int_{-1}^1 \frac{\sqrt{1-x^2}}{1+x^2}\,dx=(\sqrt 2-1)\pi$$
One way to proceed that offers a more straightforward approach is to invoke Cauchy's Integral Theorem to write
$$\begin{align} \oint_{C_{\text{Dogbone}}}\frac{\sqrt{1-z^2}}{1+z^2}\,dz&=2\int_{-1}^1 \frac{\sqrt{1-x^2}}{1+x^2}\,dx\\\\ &=\underbrace{\lim_{R\to \infty}\int_{2\pi}^0\frac{-i\sqrt{R^2e^{i2\phi}-1}}{1+R^2e^{i2\phi}}\,iRe^{i\phi}\,d\phi}_{\text{Contribution from the Residue at Infinity}}\\\\& +2\pi i \text{Res}\left(\frac{\sqrt{1-z^2}}{1+z^2}, z=i\right)\\\\ &+2\pi i \text{Res}\left(\frac{\sqrt{1-z^2}}{1+z^2},z=-i\right)\\\\ &=-2\pi+\pi \sqrt{2}+\pi\sqrt 2 \end{align}$$
Hence, we obtain the coveted result
$$\int_{-1}^1 \frac{\sqrt{1-x^2}}{1+x^2}\,dx=(\sqrt 2-1)\pi$$