$\int_0^\infty \frac{\sqrt{x}}{x^2+2x+5}\mathrm{d}x$ using Feynman's trick We are expected to solve an integral similar to $\int_0^\infty \frac{\sqrt{x}}{x^2+2x+5}\mathrm{d}x$ using contour integration, but I was wondering whether it would be possible to use the so-called Feynman's trick, i.e. differentiating under the integral sign. I tried using $$F(t) = \int_0^\infty \frac{\sqrt{x}e^{-t\sqrt{x}}}{x^2+2x+5}\mathrm{d}x$$
so that
$$F^\prime(t) = \int_0^\infty \frac{-xe^{-t\sqrt{x}}}{x^2+2x+5}\mathrm{d}x$$
Sadly, this does not work. So I'm looking for any starting point to approach this problem. I appreciate any help.
 A: I'm not sure about Feynman's trick, but there is a delightfully simple way to compute this integral using Glasser's Master Theorem. First, substitute $u=x^2$ to obtain
$$2\int_0^\infty \frac{u^2}{u^4+2u^2+5}\,du=\int_{-\infty}^\infty \frac{1}{u^2+5u^{-2}+2}\,du$$
Now, notice that $$u^2+5u^{-2}+2=\left(u-\frac{\sqrt{5}}{u}\right)^2+4\phi,$$
where $\phi=\frac{1+\sqrt{5}}{2}$.
Glasser's Master Theorem then tells us that
$$\int_{-\infty}^\infty \frac{1}{\left(u-\sqrt{5}/u\right)^2+4\phi}\,du=\int_{-\infty}^\infty \frac{1}{u^2+4\phi}\,du.$$
The last integral is just
$$\frac{\pi}{2\sqrt{\phi}}=\frac{\pi}{\sqrt{2}}\cdot\frac{1}{\sqrt{1+\sqrt{5}}}$$
A: Well, you could solve this problem using Differentiation Under the Integral Sign, however I think that it wouldn't be an easy task and probably would end up in a tricky differential equation. Instead of it, I offer you a solution that just requires some substitutions.
$$I=\int_{0}^{\infty}{\frac{\sqrt x}{x^2+2x+5}dx}\overbrace{=}^{x\rightarrow\sqrt{5t}}5^{\frac{3}{4}}\int_{0}^{\infty}{\frac{\sqrt t}{{5\ t}^2+2\sqrt5t+5}dt}$$
Let's make some rearrangements:
$$I=\color{red}{\frac{2}{\sqrt[4]{5}}\int_0^{\infty}\frac{\frac{1}{2}\frac{1}{\sqrt t}}{\left(\sqrt t-\frac{1}{\sqrt t}\right)^2+\frac{10+2\sqrt5}{5}}dt}\overbrace{=}^{t\rightarrow \frac{1}{t}}\color{blue}{\frac{2}{\sqrt[4]{5}}\int_0^{\infty}\frac{\frac{1}{2}\frac{1}{t\sqrt t}}{\left(\sqrt t-\frac{1}{\sqrt t}\right)^2+\frac{10+2\sqrt5}{5}}dt}$$
Summing the integrals red and blue:
$$2I={\frac{2}{\sqrt[4]{5}}\int_0^{\infty}\frac{\frac{1}{2}\frac{1}{\sqrt t}+\frac{1}{2}\frac{1}{t\sqrt t}}{\left(\sqrt t-\frac{1}{\sqrt t}\right)^2+\frac{10+2\sqrt5}{5}}dt}\overbrace{=}^{\sqrt t-\frac{1}{\sqrt t}=u}\frac{2}{\sqrt[4]{5}}\int_{-\infty}^{\infty}\frac{du}{u^2+\frac{10+2\sqrt5}{5}}$$
$$2I=\frac{2}{\sqrt[4]{5}}\sqrt{\frac{5}{2\left(5+\sqrt5\right)}}\left[\arctan{\left(u\sqrt{\frac{5}{2\left(5+\sqrt5\right)}}\right)}\right]_{-\infty}^\infty$$
Hence:
$$I=\frac{\pi}{\sqrt[4]{5}}\sqrt{\frac{5}{2\left(5+\sqrt5\right)}}=\frac{\pi}{2\sqrt{\phi}}$$
