Evaluating real integral using complex analysis.

Question

I'm trying to compute the following integral: $$\int_0^{\infty}\frac{\sqrt{x}}{1+x^4}dx$$

I'll not write down everything I've done, but choosing the branch cut on the positive real axes we have that:

$$\int_0^{\infty}\frac{\sqrt{x}}{1+x^4}dx=\pi i \sum_{z_i}Res(f,z_i) \qquad z_i\in\{\pm \sqrt{i},\pm\sqrt{-i}\}$$

So we have to compute four residues. My thought was changing the branch cut by putting it on the negative imaginary axes. We can do it by choosing $arg(z) \in (-\frac{\pi}{2},\frac{3\pi}{2}]$. So we have that:

$$(1+i)\int_0^{\infty}\frac{\sqrt{x}}{1+x^4}dx=2\pi i \sum_{z_i}Res(f,z_i) \qquad z_i\in\{e^{i\frac{\pi}{4}},e^{i\frac{3\pi}{4}}\}$$

By doing this, we now need to compute only two residues. But I'm really finding difficulties in computing those residues: in fact I can't obtain the result I'm expecting. Can you please show me the computation and tell me if my argument was clear and correct?

Thanks in advance.

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{\int_{0}^{\infty}{\root{x} \over 1 + x^{4}}\,\dd x = {1 \over 4}\pi\sec\pars{\pi \over 8}} = {1 \over 2}\pi\root{1 - {\root{2} \over 2}} \approx 0.8501: {\Large ?}}$. Hereafter, I'll perform an evaluation of $\ds{\oint_{\cal C}{\root{z} \over 1 + z^{4}}\,\dd z}$ where $\ds{\cal C}$ is defined in each particular case for the chosen $\ds{\root{z}}$-branch cut.

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$\ds{\Large\left.\mbox{a}\right)}$ The $\ds{\root{z}}$-branch cut is given by $$ \root{z} = \root{\verts{z}}\expo{\ic\arg\pars{z}/2}\,,\quad 0 < \arg\pars{z} < 2\pi\,\quad z \not= 0 $$ which is the OP choice. Poles are given by $\ds{p_{n} = \expo{n\pi\ic/4}\ \mbox{with}\ n = 1,3,5,7}$. Then, \begin{align} &\bbox[5px,#ffd]{\oint_{}{\root{z} \over 1 + z^{4}}\,\dd z} = 2\pi\ic\sum_{\braces{p_{n}}}{\root{p_{n}} \over 4p_{n}^{3}} \\[5mm] = &\ -\,{1 \over 2} \,\pi\ic\sum_{\braces{p_{n}}}p_{n}\root{p_{n}} = \pi\root{1 - {\root{2} \over 2}}\label{1}\tag{1} \end{align} Also, \begin{align} &\bbox[5px,#ffd]{\oint_{}{\root{z} \over 1 + z^{4}}\,\dd z} = \int_{0}^{\infty}{\root{x} \over 1 + x^{4}}\,\dd x \\[2mm] + & \require{cancel} \cancel{\mbox{integration over arc with}\ \pars{\mbox{radius}\ \to \infty}} \\[2mm] & + \int_{\infty}^{0}{\root{x}\expo{\ic\pi} \over 1 + x^{4}}\,\dd x = 2\int_{0}^{\infty}{\root{x} \over 1 + x^{4}}\,\dd x\label{2}\tag{2} \end{align} With (\ref{1}) and (\ref{2}): \begin{align} \bbox[5px,#ffd]{\int_{0}^{\infty}{\root{x} \over 1 + x^{4}}\,\dd x} & = \bbx{{1 \over 2}\pi\root{1 - {\root{2} \over 2}}} \approx 0.8501 \\ & \end{align}

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$\ds{\Large\left.\mbox{b}\right)}$ I guess the following evaluation is the simplest one because it involves just ONE pole: The integration is performed along a quarter circle in the complex plane first quadrant. The $\ds{\root{z}}$-branch cut is given by $$ \root{z} = \root{\verts{z}}\expo{\ic\arg\pars{z}/2}\,,\quad -\pi < \arg\pars{z} < \pi\,\quad z \not= 0 $$ which is the principal one. The contour encloses the pole $\ds{p = \expo{\pi\ic/4}}$. Then, \begin{align} &\bbox[5px,#ffd]{\int_{0}^{\infty} {\root{x} \over 1 + x^{4}}\dd x} \\[5mm] = &\ 2\pi\ic\,{p^{1/2} \over 4p^{3}} - \int_{\infty}^{0} {\root{y}\expo{\pi\ic/4} \over 1 + y^{4}}\, \ic\,\dd y \\[5mm] = &\ -\,{1 \over 2}\,\pi\ic\, \expo{3\pi\ic/8} + \ic\expo{\pi\ic/4}\int_{0}^{\infty} {\root{y} \over 1 + y^{4}}\,\dd y \\[5mm] \implies &\ \int_{0}^{\infty} {\root{x} \over 1 + x^{4}}\dd x = {\pars{-\pi\ic/2} \expo{3\pi\ic/8} \over 1 - \ic\expo{\pi\ic/4}} \\[5mm] = &\ \bbx{{1 \over 2}\pi\root{1 - {\root{2} \over 2}}} \approx 0.8501 \\ & \end{align}

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$\ds{\Large\left.\mbox{c}\right)}$

Ramanujan's Master Theorem: \begin{align} &\bbox[5px,#ffd]{\int_{0}^{\infty} {\root{x} \over 1 + x^{4}}\dd x} \,\,\,\stackrel{x^{4}\ \mapsto\ x}{=}\,\,\, {1 \over 4}\int_{0}^{\infty} {x^{\color{red}{3/8} - 1} \over 1 + x}\dd x \end{align} Note that $\ds{{1 \over 1 + x} = \sum_{k = 0}^{\infty}\pars{-x}^{k} = \sum_{k = 0}^{\infty}\color{red}{\Gamma\pars{1 + k}}{\pars{-x}^{k} \over k!}}$.

Then, \begin{align} &\bbox[5px,#ffd]{\int_{0}^{\infty} {\root{x} \over 1 + x^{4}}\dd x} = {1 \over 4}\bracks{\Gamma\pars{3 \over 8} \Gamma\pars{1 - {3 \over 8}}} \\[5mm] = & {1 \over 4}\,{\pi \over \sin\pars{3\pi/8}} = {1 \over 4}\,\pi\sec\pars{\pi \over 8} \\[5mm] = &\ \bbx{{1 \over 2}\pi\root{1 - {\root{2} \over 2}}} \approx 0.8501 \\ & \end{align}

Answer 2

The calcus of the residues are relatevely simple when you have simple poles.

Infact, if $z_0$ is a simple pole then $f(z) = a_{-1}(z-z_{0})^{-1}+ \sum\limits_{n \geq 0}a_n(z-z_0)^n$

So $(z-z_{0})f(z) = (z-z_{0})^{-1}+ \sum\limits_{n \geq 0}a_n(z-z_0)^n$ which implies

$$\text{Res}(f,z_{0}) = a_{-1} = \lim\limits_{z \to z_0}(z-z_{0})f(z)$$

This result to be useful when we condiser $f$ of the form $\frac{f}{q}$ with $p,q$ holomorphic function, $p(z_0) \ne 0$ and $z_0$ a simple pole of $q$ since

$$\text{Res}(f,z_{0})= a_{-1} = \lim\limits_{z \to z_0}(z-z_{0})\frac{p(z)}{q(z)} = \frac{p(z_0)}{q'(z_0)}$$

In general :

For higher order poles a strategy could be : If $f$ has a pole of order $k$ in $z_0$, $g(z) = (z-z_0)^k f(k)$ extends to an holomorphic function in $z_0$ (I'm gonna call it improperly by $g$ as well)

With this setting $$f(z) = a_{-k}(z-z_0)^k + \cdots + a_{-1}(z-z_0)^{-1} + \sum\limits_{n \geq 0}a_n(z-z_0)^n$$

$$g(z) = a_{-k} + \cdots + a_{-1}(z-z_0)^{k-1} + \sum\limits_{n \geq 0}a_n(z-z_0)^{n+k}$$

So $a_{-1}$ is the coefficient of $(z-z_0)^{k-1}$ in the expansion of $g$ which is holomorphic. Knowing that $a_{n} = \frac{f^{(n)}(z_0)}{n!}$ we have $$\text{Res}(f,z_{0}) = a_{-1} = \frac{g^{(k-1)}(z_0)}{(k-1)!}$$

Hope this helps with your calculations.

Answer 3

Under $x^4\to x$, $$\int_0^{\infty}\frac{\sqrt{x}}{1+x^4}dx=\frac14\int_0^\infty\frac{1}{x^{5/8}(1+x)}dx. $$ Let $$ f(z)=\frac{1}{z^{5/8}(1+z)}. $$ Let $C_r, C_R$ be circles at $0$ cut from $r$ to $R$, respectively, and $C_1, C_2$ be the top and bottom parts of the segment from $r$ to $R$. Then, for big $R>0$ and small $r>0$, $$ \int_{C_R}f(z)dz+\int_{C_r^-}f(z)dz+\int_0^{R}f(x)dx-\int_0^{R}f(xe^{2\pi i})dx=2\pi i\text{Res}(f,z=-1). $$ Clearly $$ \bigg|\int_{C_R}f(z)dz\bigg|\le\frac{1}{R^{5/8}(R-1)}2\pi R=\frac{2\pi R^{3/8}}{R-1}, \bigg|\int_{C_r^-}f(z)dz\bigg|\le\frac{1}{r^{5/8}(1-r)}2\pi r=\frac{2\pi r^{3/8}}{1-r} $$ and $$ \int_0^{R}f(xe^{2\pi i})dx=e^{-5\pi i/4}\int_0^\infty f(x)dx, \text{Re}(f,z=-1)=e^{-5\pi i/8}. $$ So letting $R\to\infty, r\to 0^+$, one has $$ (1+e^{-5\pi i/4})\int_0^\infty f(x)dx=2\pi i e^{-5\pi i/8} $$ or $$ \int_0^\infty f(x)dx=\frac{2\pi i e^{-5\pi i/8}}{1+e^{-5\pi i/4}}=\frac{\pi}{\cos(\pi/8)}. $$ Thus $$\int_0^{\infty}\frac{\sqrt{x}}{1+x^4}dx=\frac14\int_0^\infty\frac{1}{x^{5/8}(1+x)}dx=\frac{\pi}{4\cos(\pi/8)}. $$