Evaluate $\int_0^{\infty}\frac{\sqrt{x}}{x^2+1}dx$ without Complex Analysis I found  this question on a Complex Analysis Qualifying Exam:
How would you evaluate: $$\int_0^{\infty}\frac{\sqrt{x}}{x^2+1}dx$$
I am interested in all methods including Complex Analysis, I'm just less familiar with it. 
For instance, since the integral is from $0$ to $\infty$ do we only consider the $x-i$ factor in $x^2+1=(x-i)(x+i)$ since $i$ would be the only root in the upper half plane?

Without complex analysis? Okay... 
Let $$I=\int_0^\infty\frac{\sqrt{x}}{x^2+1}dx$$ 
Enforce $x:=y^2\implies dx=2y \space dy$. So $$I=\int_0^\infty\frac{\sqrt{x}}{x^2+1}dx=2\int_0^\infty \frac{y^2}{y^4+1}dy=2\int_0^\infty\frac{y^2}{y^2\left(y^2+\frac{1}{y^2}\right)}dy$$ $$=\int_0^\infty \frac{1-\frac{1}{y^2}+1+\frac{1}{y^2}}{y^2+\frac{1}{y^2}}dy=\int_0^\infty\frac{1-1/y^2}{(y+1/y)^2-2}dy+\int_0^\infty\frac{1+1/y^2}{(y-1/y)^2+2}dy\tag1$$
For the first of the two integrals  let $t=y+1/y\implies dt=(1-1/y^2)\space dy$. For the second, let $u=y-1/y\implies du=(1+1/y^2)\space dy$. Hence $(1)$ becomes: $$I=\int_{\infty}^\infty\frac{dt}{t^2-2}+\int_{-\infty}^\infty\frac{du}{u^2+2}=0+\int_{-\infty}^\infty\frac{du}{u^2+2}=2\int_0^\infty\frac{du}{2\left(\left(\frac{u}{\sqrt{2}}\right)^2+1\right)}\tag2$$
Finally with the substitution $z=\frac{u}{\sqrt{2}}\implies dz=\frac{du}{\sqrt{2}}$ $(2)$ becomes $$I=\sqrt{2}\int_0^\infty\frac{dz}{z^2+1}=\sqrt{2}\left(\arctan(z)\big|^\infty_0\right)=\sqrt{2}\left(\frac{\pi}{2}\right)=\frac{\pi}{\sqrt{2}}$$
Thus $$\boxed{I=\frac{\pi}{\sqrt{2}}}$$
