complex integrationcomplex-analysisdefinite integralsintegration
When calculating integrals like $$\int_0^\infty \frac{x^\alpha}{1 + x^2}dx$$ for $\alpha \in (-1,1 )$, it is convenient to take the branch cut of the integrand along the positive real axis and then use the keyhole contour.
I was wondering is there a way to use the principal branch cut, which runs along the negative real axis, to calculate this kind of integrals? I'm assuming some trivial manipulation of the integrand for $x>0$ would do the trick, but I fail to see it.
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
Using the principal branch, we can write the integral $\oint_C \frac{z^a}{z^2+1}\,dz$, where $C$ is comprised of (i) the real line segment from $-R$ to $R$ and (ii) the semi-circle in the upper-half plane, centered at the origin and with radius $R$, as
$$\begin{align} \oint_C \frac{z^a}{z^2+1}\,dz&=\int_{-R}^0 \frac{x^a}{x^2+1}\,dx+\int_0^R \frac{x^a}{x^2+1}\,dx+\int_0^\pi \frac{(Re^{i\phi})^a}{(Re^{i\phi}))^2+1}\,(iRe^{i\phi}))\,d\phi\\\\ &=(1+e^{i\pi a})\int_0^R \frac{x^a}{x^2+1}\,dx+\int_0^\pi \frac{(Re^{i\phi})^a}{(Re^{i\phi}))^2+1}\,(iRe^{i\phi}))\,d\phi\tag1 \end{align}$$
As $R\to \infty$ the second integral on the right-hand side of $(1)$ approaches $0$. Hence, taking this limit and invoking the reside theorem we find that
$$\begin{align} \int_0^R \frac{x^a}{x^2+1}\,dx&=\frac1{1+e^{i\pi a}}\,(2\pi i) \text{Res}\left(\frac{z^a}{z^2+1}, z=i\right)\\\\ &=\frac{\pi e^{i\pi a/2}}{1+e^{i\pi a}}\\\\ &=\frac{\pi}{2\cos(\pi a/2)} \end{align}$$