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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.

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    $\begingroup$ Substitute $u^2=x$ then you have a rational function in $u$. $\endgroup$ Dec 7 '19 at 4:41
  • $\begingroup$ Why would you look to use Feynman's trick for such an integral? $\endgroup$
    – Mark Viola
    Dec 7 '19 at 4:48
  • $\begingroup$ @MarkViola I need to solve $\int_0^\infty \frac{x^{1/3}}{x^2+2x+5}\mathrm{d}x$. This was a bad example. $\endgroup$ Dec 7 '19 at 4:49
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    $\begingroup$ It's not a great idea to introduce $e^{-t\sqrt{x}}$ into the integral since it doesn't work well with rational functions. Mathematica gave a closed form of F' involving exponential integrals. $\endgroup$
    – Edward H
    Dec 7 '19 at 5:16
  • $\begingroup$ It is basically this: en.wikipedia.org/wiki/… $\endgroup$ Jan 17 '20 at 20:10
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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 $$\frac{\pi}{2\sqrt{\phi}}=\frac{\pi}{\sqrt{2}}\cdot\frac{1}{\sqrt{1+\sqrt{5}}}$$

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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: $$I=\frac{\pi}{\sqrt[4]{5}}\sqrt{\frac{5}{2\left(5+\sqrt5\right)}}=\frac{\pi}{2\sqrt{\phi}}$$

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