Evaluating real integral using complex analysis.

$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},} newcommand{braces}[1]{leftlbrace,{#1},rightrbrace} newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack} newcommand{dd}{mathrm{d}} newcommand{ds}[1]{displaystyle{#1}} newcommand{expo}[1]{,mathrm{e}^{#1},} newcommand{ic}{mathrm{i}} newcommand{mc}[1]{mathcal{#1}} newcommand{mrm}[1]{mathrm{#1}} newcommand{on}[1]{operatorname{#1}} newcommand{pars}[1]{left(,{#1},right)} newcommand{partiald}[3][]{frac{partial^{#1} #2}{partial #3^{#1}}} newcommand{root}[2][]{,sqrt[#1]{,{#2},},} newcommand{totald}[3][]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}} newcommand{verts}[1]{leftvert,{#1},rightvert}$ $ds{bbox[5px,#ffd]{int_{0}^{infty}{root{x} over 1 + x^{4}},dd x = {1 over 4}pisecpars{pi over 8}} = {1 over 2}piroot{1 - {root{2} over 2}} approx 0.8501: {Large ?}}$. 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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$ds{Largeleft.mbox{a}right)}$ The $ds{root{z}}$-branch cut is given by $$ root{z} = root{verts{z}}expo{icargpars{z}/2},,quad 0 < argpars{z} < 2pi,quad z not= 0 $$ which is the OP choice. Poles are given by $ds{p_{n} = expo{npiic/4} mbox{with} n = 1,3,5,7}$. Then, begin{align} &bbox[5px,#ffd]{oint_{}{root{z} over 1 + z^{4}},dd z} = 2piicsum_{braces{p_{n}}}{root{p_{n}} over 4p_{n}^{3}} \[5mm] = & -,{1 over 2} ,piicsum_{braces{p_{n}}}p_{n}root{p_{n}} = piroot{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{icpi} over 1 + x^{4}},dd x = 2int_{0}^{infty}{root{x} over 1 + x^{4}},dd xlabel{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}piroot{1 - {root{2} over 2}}} approx 0.8501 \ & end{align}

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$ds{Largeleft.mbox{b}right)}$ 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 $$ root{z} = root{verts{z}}expo{icargpars{z}/2},,quad -pi < argpars{z} < pi,quad z not= 0 $$ which is the principal one. The contour encloses the pole $ds{p = expo{piic/4}}$. Then, begin{align} &bbox[5px,#ffd]{int_{0}^{infty} {root{x} over 1 + x^{4}}dd x} \[5mm] = & 2piic,{p^{1/2} over 4p^{3}} - int_{infty}^{0} {root{y}expo{piic/4} over 1 + y^{4}}, ic,dd y \[5mm] = & -,{1 over 2},piic, expo{3piic/8} + icexpo{piic/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{-piic/2} expo{3piic/8} over 1 - icexpo{piic/4}} \[5mm] = & bbx{{1 over 2}piroot{1 - {root{2} over 2}}} approx 0.8501 \ & end{align}

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$ds{Largeleft.mbox{c}right)}$

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 $ds{{1 over 1 + x} = sum_{k = 0}^{infty}pars{-x}^{k} = sum_{k = 0}^{infty}color{red}{Gammapars{1 + k}}{pars{-x}^{k} over k!}}$.

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

Under $x^4to x$, $$int_0^{infty}frac{sqrt{x}}{1+x^4}dx=frac14int_0^inftyfrac{1}{x^{5/8}(1+x)}dx. $$ 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 $r>0$, $$ int_{C_R}f(z)dz+int_{C_r^-}f(z)dz+int_0^{R}f(x)dx-int_0^{R}f(xe^{2pi i})dx=2pi itext{Res}(f,z=-1). $$ Clearly $$ bigg|int_{C_R}f(z)dzbigg|lefrac{1}{R^{5/8}(R-1)}2pi R=frac{2pi R^{3/8}}{R-1}, bigg|int_{C_r^-}f(z)dzbigg|lefrac{1}{r^{5/8}(1-r)}2pi r=frac{2pi r^{3/8}}{1-r} $$ and $$ int_0^{R}f(xe^{2pi i})dx=e^{-5pi i/4}int_0^infty f(x)dx, text{Re}(f,z=-1)=e^{-5pi i/8}. $$ So letting $Rtoinfty, rto 0^+$, one has $$ (1+e^{-5pi i/4})int_0^infty f(x)dx=2pi i e^{-5pi i/8} $$ or $$ int_0^infty f(x)dx=frac{2pi i e^{-5pi i/8}}{1+e^{-5pi i/4}}=frac{pi}{cos(pi/8)}. $$ Thus $$int_0^{infty}frac{sqrt{x}}{1+x^4}dx=frac14int_0^inftyfrac{1}{x^{5/8}(1+x)}dx=frac{pi}{4cos(pi/8)}. $$

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

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

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

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

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

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

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 $$f(z) = a_{-k}(z-z_0)^k + cdots + a_{-1}(z-z_0)^{-1} + sumlimits_{n geq 0}a_n(z-z_0)^n$$

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

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.