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I need to evaluate the integral $$\int_0^{\infty}\frac{dx}{\sqrt[4]{x}(1+x^2)}$$

I've been given the hint to use the keyhole contour. So I would first choose the principal branch of $\sqrt[4]{\cdot}$, then I have the "keyhole" around $0$, giving me

$$2\pi i\left(\text{Res}_if+\text{Res}_{-i}f\right)= \\ \int_{\gamma_{R,\epsilon}}\frac{dz}{\sqrt[4]{z}(1+z^2)}=\int_{\gamma_R}f(z)dz-\int_{\gamma_\epsilon}f(z)dz+\int^R_{\epsilon} f(z)dz - \int^R_{\epsilon} f(-z)dz$$

($f$ is the integrand) Then taking $R \rightarrow \infty$ and $\epsilon \rightarrow 0$, should give me my result. But I happened to check out the integral in Wolfram Alpha and it gives the integral as $\frac{1}{2}\pi\sec\left(\frac{\pi}{8}\right)$, which is not what I get. Can I get some help? I'm sure I've gone wrong somewhere, and I'm pretty new at using these arguments, so help or insights will be nice.

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  • $\begingroup$ Could you please show what you get? $\sec \frac\pi8$ can be simplified into radical form. $\endgroup$ – Kemono Chen Nov 7 '18 at 3:14
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Keeping it simple we introduce

$$f(z) = \exp(-(1/4)\mathrm{Log}(z)) \frac{1}{1+z^2}$$

with the branch cut of the logarithm on the positive real axis and argument from $0$ to $2\pi.$ The slot of the keyhole rests on the positive real axis and the contour is traversed counter-clockwise. Let the segment above the real axis be $\Gamma_1,$ the large circle $\Gamma_2$, the segment below the positive real axis $\Gamma_3$ and the small circle around the origin $\Gamma_4.$

We get for $\Gamma_1$ in the limit

$$J = \int_0^\infty \frac{1}{\sqrt[4]{x}} \frac{1}{1+x^2} \; dx,$$

i.e. the target integral. The contribution from the circlular components vanishes in the limit. We get below the cut on $\Gamma_3$ in the limit

$$\exp(-(1/4)2\pi i) \int_\infty^0 \frac{1}{\sqrt[4]{x}} \frac{1}{1+x^2} \; dx \\= - \exp(-(1/2)\pi i) \int_0^\infty \frac{1}{\sqrt[4]{x}} \frac{1}{1+x^2} \; dx = iJ.$$

We have for the first residue at the pole $z=i$

$$\left.\exp((-1/4)\mathrm{Log}(z))\frac{1}{z+i}\right|_{z=i} = \exp((-1/4)\pi i/2 )\frac{1}{2i}$$

and for the second one

$$\left.\exp((-1/4)\mathrm{Log}(z))\frac{1}{z-i}\right|_{z=-i} = -\exp((-1/4)3\pi i/2)\frac{1}{2i}.$$

Collecting everything we have

$$(1+i) J = 2\pi i \frac{1}{2i} (\exp(-\pi i/8) - \exp(-3\pi i/8)).$$

This is

$$J = \pi (\exp(-\pi i/8) - \exp(-3\pi i/8)) \frac{1}{\sqrt{2}} \exp(-\pi i /4) \\ = \frac{\sqrt{2}}{2} \pi (\exp(-3\pi i/8) - \exp(-5\pi i/8)) \\ = \frac{\sqrt{2}}{2} \pi \exp(-4\pi i/8) (\exp(\pi i/8) - \exp(-\pi i/8)) \\ = \frac{\sqrt{2}}{2} \pi \exp(-\pi i/2) \times 2i \sin(\pi/8).$$

The end result is

$$\bbox[5px,border:2px solid #00A000]{ \sqrt{2} \times \pi \times \sin(\pi/8).}$$

Remark. As per the contribution from the circles vanishing, we get for the large circle $\Gamma_2$ $\lim_{R\to\infty} 2\pi R / R^{1/4} / R^2 = 0$ and for the small one $\Gamma_4$ $\lim_{\epsilon\to 0} 2\pi \epsilon / \epsilon^{1/4} / 1 = 0.$

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Sorry but it could be solved much easily with beta and gamma functions.

$$\int_0^{\infty} \frac {dx}{\sqrt[4] x (1+x^2)}=\frac 12 B\left(\frac 38,\frac 58\right)=\frac 12\Gamma(3/8)\Gamma(5/8)=\frac 12\pi\csc (3\pi/8)=\frac 12\pi\sec(\pi/8)$$

PS: Please don't downvote just because I used a different method. I just thought to introduce a little bit of advanced integration techniques other than used in complex analysis

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  • $\begingroup$ I really appreciate your answer, to be frank I'm not aware of these techniques. Could you provide me with any resources where I can read up on them? Are they available in Alfhors' or Lang's Complex Analysis books? $\endgroup$ – Naweed G. Seldon Nov 8 '18 at 3:31
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Yet another way: we may remove the branch point at the origin by setting $x=z^4$, leading to $$ I = \int_{0}^{+\infty}\frac{dx}{\sqrt[4]{x}(1+x^2)}=\int_{0}^{+\infty}\frac{4z^2\,dz}{1+z^8}=\int_{0}^{1}\frac{4z^2\,dz}{1+z^8}+\int_{0}^{1}\frac{4z^4}{z^8+1}\,dz$$ then to $$ I = 4\int_{0}^{1}\frac{(z^2+z^4)(1-z^8)}{1-z^{16}}\,dz=4\sum_{n\geq 0}\left[\tfrac{1}{16n+3}+\tfrac{1}{16n+5}-\tfrac{1}{16n+11}-\tfrac{1}{16n+13}\right].$$ Now the Dirichlet $L$-series appearing in the RHS can be computed by recalling that $$ \sum_{n\geq 0}\left[\frac{1}{an+b}-\frac{1}{an+(a-b)}\right]=\frac{\pi}{a}\cot\left(\frac{\pi b}{a}\right) $$ holds by the reflection formula for the $\psi$ function / Herglotz trick. We get $$ I = \frac{\pi}{4}\left[\cot\left(\frac{3\pi}{16}\right)+\cot\left(\frac{5\pi}{16}\right)\right]=\color{red}{\frac{\pi}{\sqrt{2+\sqrt{2}}}}. $$

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\int_{0}^{\infty}{\dd x \over \root[{\large 4}]{x}\pars{1 + x^{2}}}} \,\,\,\stackrel{x^{2}\ \mapsto\ x}{=}\,\,\, {1 \over 2}\int_{0}^{\infty}{x^{-5/8} \over 1 + x}\,\dd x \\[5mm] \stackrel{x + 1\ \mapsto\ x}{=}\,\,\,&\ {1 \over 2}\int_{1}^{\infty}{\pars{x - 1}^{-5/8} \over x}\,\dd x \,\,\,\stackrel{x\ \mapsto\ 1/x}{=}\,\,\, {1 \over 2}\int_{1}^{0}{\pars{1/x - 1}^{-5/8} \over 1/x}\, \pars{-\,{\dd x \over x^{2}}} \\[5mm] = &\ {1 \over 2}\int_{0}^{1}x^{-3/8}\,\pars{1 - x}^{-5/8}\,\dd x = {1 \over 2}\,{\Gamma\pars{5/8}\Gamma\pars{3/8} \over \Gamma\pars{1}} = {1 \over 2}\,{\pi \over \sin\pars{3\pi/8}} \\[5mm] = &\ \bbx{{1 \over 2}\pi\sec\pars{\pi \over 8}} \approx 1.7002 \end{align}

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