Indented Path Integration

Question

The goal is to show that $$\int_0^\infty \frac{x^{1/3}\log(x)}{x^2 + 1}dx = \frac{\pi^2}{6}$$ and that $$\int_0^\infty \frac{x^{1/3}}{x^2 + 1}dx = \frac{\pi}{\sqrt{3}}.$$

So, we start with the function $$f(z) = \frac{z^{1/3}\log(z)}{z^2 + 1}.$$

Let $c_r$ be the upper semicircle with radius r such that $r< 1$ and let $c_R$ be the upper semicircle with radius R such that $R>1$. Let $L_1$ be the interval $(r,R)$ on the reals and let $L_2$ be the interval $(-R,r)$ on the reals.

Then $$\int_{c_r} f + \int_{c_R} f + \int_{L_1} f + \int_{L_2} f$$ taken counterclockwise is equal to $$2\pi i\text{ res}(f,i) = -\frac{\pi^2 e^{\frac{\pi i}{6}}}{2}$$

The integrals along $c_r$ and $c_R$ should go to $0$ as $r$ goes to $0$ and $R$ goes to infinity.

So: $$\int_{L_1} + \int_{L_2}= \int_0^{\infty}\frac{x^{1/3}\log(x)}{x^2 + 1}dx - \int_{0}^{\infty} \frac{x^{1/3}e^{i\pi/3}(ln(x) + i\pi) }{x^2 + 1}dx$$ where x is the polar radius.

Let $A = \int_0^\infty \frac{x^{1/3}\log(x)}{x^2 + 1}dx$ and let $B = \int_0^\infty \frac{x^{1/3}}{x^2 + 1}dx $.

Then we have $A*(1-e^{\frac{i\pi}{3}}) - \pi i e^{\frac{i\pi}{3}}B = $-$\frac{\pi^2 e^{\frac{\pi i}{6}}}{2}$.

This doesn't mesh with the given values for $A$ and $B$, so I'm wondering where I went wrong. Thanks in advance.

Answer 1

In this case, I would use a keyhole contour about the positive real axis. Consider the contour integral

$$\oint_C dz \frac{z^{1/3}}{1+z^2} \log{z}$$

where $C$ is the keyhole contour. As it is clear that the integral about the circular arcs, large and small, vanish as their respective radii go to infinity and zero, the contour integral is equal to

$$\int_0^{\infty} dx \frac{x^{1/3}}{1+x^2} \log{x} - e^{i 2 \pi/3} \int_0^{\infty} dx \frac{x^{1/3}}{1+x^2} (\log{x}+i 2 \pi) $$

which is equal to

$$\left (1-e^{i 2 \pi/3} \right )\int_0^{\infty} dx \frac{x^{1/3}}{1+x^2} \log{x} - i 2 \pi e^{i 2 \pi/3} \int_0^{\infty} dx \frac{x^{1/3}}{1+x^2}$$

By the residue theorem, the contour integral is $i 2 \pi$ times the sum of the residues at the poles $z=e^{i \pi/2}$ and $z=e^{i 3 \pi/2}$, or

$$i 2 \pi \frac{e^{i \pi/6}}{2 i} i \frac{\pi}{2} + i 2 \pi \frac{e^{i \pi/2}}{-2 i} i \frac{3 \pi}{2} = \frac{5 \pi^2}{4} + i \frac{\sqrt{3} \pi^2}{4}$$

As you did above, let $A$ be the first integral and $B$ be the second. Then

$$\left (1-e^{i 2 \pi/3} \right ) A - i 2 \pi e^{i 2 \pi/3} B = \frac{5 \pi^2}{4} + i \frac{\sqrt{3} \pi^2}{4}$$

So we equate real and imaginary parts. Rearranging, we get two equations and two unknowns:

$$\frac{3}{2} A + \sqrt{3} \pi B = \frac{5 \pi^2}{4} $$ $$-\frac{\sqrt{3}}{2} A + \pi B = \frac{\sqrt{3} \pi^2}{4}$$

From this, you may easily verify that

$$A = \int_0^{\infty}dx \frac{x^{1/3}}{1+x^2} \log{x} = \frac{\pi^2}{6}$$

$$B = \int_0^{\infty}dx \frac{x^{1/3}}{1+x^2} = \frac{\pi}{\sqrt{3}}$$

Answer 2

Indented contours pick up half a residue. So the integral along $c_r$ is not zero.

Answer 3

First, make the substitution $t=\displaystyle\frac1{1+x^2}$ , and rewrite $I_1=\displaystyle\frac14\int_0^1t^{^{-\frac16-\frac12}}\cdot(1-t)^{^{\frac16-\frac12}}\ln{1-t\over t}dt$ and $I_2=\displaystyle\frac12\int_0^1t^{^{-\frac16-\frac12}}\cdot(1-t)^{^{\frac16-\frac12}}dt$ . Then recognize the expression of Euler's Beta function in each, and use the fact that $\displaystyle\ln\frac ab=\ln a-\ln b$ , in order to break up the first integral into two parts, and lastly employ the Taylor series expansion for the natural logarithm, $\displaystyle\ln(1-t)=-\sum_{n=1}^\infty\frac{t^n}n$ to deliver the final blow. Also, for the second integral, you will have to make use of Euler's reflection formula for the Gamma function, $\displaystyle\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin{(\pi z)}}$ .

Answer 4

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ Let's consider the integral $\ds{\int_{0}^{\infty}{x^{\mu} \over x^{2} + 1}\,\dd x}$ which can be straightforward evaluated in a 'PACMAN' style contour in the complex plane:

\begin{align} &2\pi\ic\pars{{\expo{\ic\pi\mu/2} \over 2\ic} + {\expo{-\ic\pi\mu/2} \over -2\ic}} = \color{#ff0000}{\large{2\pi\ic\ \sin\pars{\pi\mu \over 2}}} \\[3mm]&=\int_{-\infty}^{0}{\pars{-x}^{\mu}\expo{\ic\pi\mu} \over x^{2} + 1}\,\dd x + \int^{-\infty}_{0}{\pars{-x}^{\mu}\expo{-\ic\pi\mu} \over x^{2} + 1}\,\dd x = \pars{\expo{-\ic\pi\mu} - \expo{\ic\pi\mu}} \int^{-\infty}_{0}{\pars{-x}^{\mu} \over x^{2} + 1}\,\dd x \\[3mm]&=\color{#ff0000}{\large% -2\ic\sin\pars{\pi\mu}\bracks{-\int^{\infty}_{0}{x^{\mu} \over x^{2} + 1}\,\dd x}} \end{align}

$$ \int^{\infty}_{0}{x^{\mu} \over x^{2} + 1}\,\dd x = {\pi \over 2}\,{1 \over \cos\pars{\pi\mu/2}}\tag{1} $$

1. $$ \color{#0000ff}{\large\int^{\infty}_{0}{x^{1/3} \over x^{2} + 1}\,\dd x} = {\pi \over 2}\,{1 \over \cos\pars{\pi/6}} = {\pi \over 2}{1 \over \root{3}/2} = \color{#0000ff}{\large{\pi \over \root{3}}} $$
2. We take the derivative, respect $\mu$, in both members of the expression $\pars{1}$: $$ \int^{\infty}_{0}{x^{\mu}\ln\pars{x} \over x^{2} + 1}\,\dd x = {\pi^{2} \over 4}\,{\sin\pars{\pi\mu/2} \over \cos^{2}\pars{\pi\mu/2}} $$ Set $\mu = 1/3$ in both members of this expression: $$ \color{#0000ff}{\large\int^{\infty}_{0}{x^{1/3}\ln\pars{x} \over x^{2} + 1} \,\dd x} = {\pi^{2} \over 4}\,{\sin\pars{\pi/6} \over \cos^{2}\pars{\pi/6}} = {\pi^{2} \over 4}\,{1/2 \over \pars{\root{3}/2}^{2}} =\color{#0000ff}{\large{\pi^{2} \over 6}} $$

Answer 5

I know it specifically says not to ask for clarification in an answer but I am confused on one specific point in the answer given by Ron Gordon to the original question. In your second integral, where does the $$e^{i 2 \pi / 3}$$ come from? I understand that the arg there is taken to be $2 \pi$ but that doesn't explain why that is needed.