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

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!

• How do you pass from $dx$ to $du$? – dan_fulea Oct 22 '18 at 18:08
• $dx=2udu$ you are missing the factor $2$ after the substitution. – N74 Oct 22 '18 at 18:08

If $$x=u^2$$, then $$\mathrm dx=2u\,\mathrm du$$. Here's the factor $$2$$ that you missed.
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$$\int\limits_0^{\infty}\mathrm dx\,\frac {\sqrt x}{x^2+4}\color{blue}{=\frac {\pi}2}$$