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Problem: Calculate $$\int_0^\infty\frac{\sqrt{x}}{x^2+4}dx$$

I did some calculations but for some reason I end up with half of the value it's supposed to have. Maybe someone can find my error:

First I substitute $u:=\sqrt x$, this yields the integral $$I=\int_0^\infty\frac{u^2}{u^4+4}du=\frac{1}{2}\int_{\mathbb R}\frac{u^2}{u^4+4}du$$

Now I integrate the function $f(z)=\frac{z^2}{z^4+4}$ along the path $\gamma_1\circ\gamma_2$ where $$\gamma_1:\ [-R,R]\to\mathbb C,\ t\mapsto t$$ and $$\gamma_2: [0,\pi]\to\mathbb C,\ t\mapsto Re^{it}$$ For the integral along $\gamma_2$ I obtain via the standard estimation that $\int_{\gamma_2}f(z)dz\to 0$ as $R\to\infty$ so we have $$2I=\lim_{R\to\infty}\int_{\gamma_1}f(z)dz=\lim_{R\to\infty}\oint_{\gamma_1\circ\gamma_2}f(z)dz$$ The rest is just the residue theorem: $f$ has 4 poles of order 1 at $\pm1\pm i$ where only $\pm 1+i$ are in the half-circle created by $\gamma_1\circ\gamma_2$. I let wolfram alpha do the work and obtain $$Res(f,1+i)=\frac{1}{8}(1-i)$$ $$Res(f,-1+i)=\frac{1}{8}(-1-i)$$ so we have $$2I=2\pi i(Res(f,1+i)+Res(f,-1+i)=\frac{\pi}{2}$$ but if I type in the integral at the beginning it says $I=\frac{\pi}{2}$, so somewhere I must have lost a factor of 2 but I can't find it. Maybe this is a duplicate but I am really eager to find the mistake I did in these calculations.

Thanks!

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  • $\begingroup$ How do you pass from $dx$ to $du$? $\endgroup$ – dan_fulea Oct 22 '18 at 18:08
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    $\begingroup$ $dx=2udu$ you are missing the factor $2$ after the substitution. $\endgroup$ – N74 Oct 22 '18 at 18:08
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If $x=u^2$, then $\mathrm dx=2u\,\mathrm du$. Here's the factor $2$ that you missed.

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  • $\begingroup$ Thanks! Jesus, I've been looking at my notes for a good 30 minutes and couldn't see it. $\endgroup$ – Buh Oct 22 '18 at 19:56
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Here's another way to calculate the integral using real analysis. Make the substitution $x\mapsto 2x$ so that$$\mathfrak{I}=\frac 1{\sqrt2}\int\limits_0^{\infty}\mathrm dx\,\frac {\sqrt x}{x^2+1}$$Now let $x\mapsto\tan x$ so that$$\mathfrak{I}=\frac 1{\sqrt2}\int\limits_0^{\pi/2}\mathrm dx\,\sin^{1/2}x\cos^{-1/2}x=\frac 1{2\sqrt2}\operatorname{B}\left(\frac 34,\frac 14\right)$$Through the reflection formula for the gamma function$$\Gamma\left(\frac 14\right)\Gamma\left(\frac 34\right)=\pi\sqrt2$$Therefore$$\int\limits_0^{\infty}\mathrm dx\,\frac {\sqrt x}{x^2+4}\color{blue}{=\frac {\pi}2}$$

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