A calculus problem by Refaat M. Sayed

$\large I = \int_0^{\infty} \frac{\sqrt{x}}{x^2 + \sqrt{3}x + 1} dx$

Make $\sqrt{x} = u$ Then $\frac{1}{2\sqrt{x}}dx = du$ or $dx = 2udu$

$\large I = 2 \int_0^{\infty} \frac{u^2}{u^4 + \sqrt{3}u^2 + 1} du$

$\large I = 2 \int_0^{\infty} \frac{u^2}{(u^2 + \frac{\sqrt{3} + j}{2})\cdot (u^2 + \frac{\sqrt{3} - j}{2})} du$

Being $j$ the imaginary unit.

$\large I = 2 \int_0^{\infty} u^2 \Big( \frac{j}{u^2 + e^{j\pi/6}} - \frac{j}{u^2 + e^{-j\pi/6}} \Big) du$

$\large I = 2j \int_0^{\infty} \Big( \frac{u^2}{u^2 + e^{j\pi/6}} - \frac{u^2}{u^2 + e^{-j\pi/6}} \Big) du$

$\large I = 2j \Big[ \Big(u - e^{j\pi/12}\cdot atan(u\cdot e^{-j\pi/12}) \Big) \Big|_0^{\infty} - \Big(u - e^{-j\pi/12}\cdot atan(u\cdot e^{j\pi/12}) \Big) \Big|_0^{\infty} \Big]$

$\large I = 2j \cdot (e^{-j\pi/12} \cdot \frac{\pi}{2} - e^{j\pi/12} \cdot \frac{\pi}{2})$

$\large I = 2 \pi sin(\frac{\pi}{12})$

$\large I = \frac{\pi sin(\frac{\pi}{12})}{\frac{1}{2} \cdot 1}$

$\large I = \frac{\pi sin(\frac{\pi/6}{2})}{\sin(\pi/6) \cdot sin (3 \cdot \pi/6)}$

Hence $A = \pi /6$ or, in degrees, $A = 30^o$

The powers of $30$ divided by $7$ generate a remainder in the pattern $\{2, 4, 1, 2, 4, 1...\}$ , i.e., $2$ if the power is a multiple of $3$ plus $1$ , $4$ if the power is a multiple of $3$ plus $2$ and $1$ if the power is a multiple of $3$ .

Since $30$ is a multiple of $3$ , the remainder is $\color{#3D99F6} \fbox{1}$

Relevant wiki: Cauchy Integral Formula

Consider the complex form of $\displaystyle I = \int_0^\infty f(x) \ dx = \int_0^\infty \frac {\sqrt x}{x^2+\sqrt 3 x + 1} dx$ ,

$\begin{aligned} J & = \int_C \frac {\color{#3D99F6}\sqrt z}{z^2+\sqrt 3 z + 1} dz & \small \color{#3D99F6} \text{As }\sqrt z \text{ has a branch cut, we use the keyhole contour.} \\ & = \int_\epsilon^R f(z) \ dz + {\color{#3D99F6} \int_\Gamma f(z) \ dz} + \int^\epsilon_R f(z) \ dz + {\color{#3D99F6} \int_\gamma f(z) \ dz} & \small \color{#3D99F6} \text{Note: } \lim_{R \to \infty} \int_\Gamma f(z) \ dz = 0, \ \lim_{\epsilon \to 0} \int_\gamma f(z) \ dz = 0 \\ & = \int_\epsilon^R \frac {\sqrt z}{z^2+\sqrt 3 z + 1} \ dz + {\color{#3D99F6} \int^\epsilon_R \frac {\sqrt z}{z^2+\sqrt 3 z + 1} \ dz} & \small \color{#3D99F6} \text{See note: }\int^\epsilon_R f(z) \ dz = \int_\epsilon^R f(z) \ dz \\ & = 2 \int_0^\infty \frac {\sqrt x}{x^2+\sqrt 3 x + 1} \ dz \end{aligned}$

$\implies J = 2I$ and then we have:

$\begin{aligned} I & = \frac 12 \int_C \frac {\sqrt z}{z^2+\sqrt 3 z + 1} dz \\ & = \frac 12 \int_C \frac {\sqrt z}{\left(z+\frac {\sqrt 3}2 - \frac i2 \right)\left(z+\frac {\sqrt 3}2 + \frac i2 \right)} dz \\ & = \frac 12 \int_C \frac {\sqrt z}{\left(z - {\color{#3D99F6}e^{\frac {5 \pi}6i}} \right)\left(z - {\color{#3D99F6}e^{\frac {7 \pi}6i}} \right)} dz & \small \color{#3D99F6} \text{Two poles within the contour }z = e^{\frac {5 \pi}6i},\ e^{\frac {7 \pi}6i} \\ & = \frac 1{2i} \int_C \left( \frac {\sqrt z}{z - e^{\frac {5 \pi}6i}} + \frac {\sqrt z}{z - e^{\frac {7 \pi}6i}} \right) dz & \small \color{#3D99F6} \text{By Cauchy integral formula} \\ & = \frac {2\pi i}{2i} \left(e^{\frac {5 \pi}{12}i} - e^{\frac {7 \pi}{12}i} \right) \\ & = \pi \left(\cos \frac {5 \pi}{12} + i \sin \frac {5 \pi}{12} - \cos \frac {7 \pi}{12} - i\sin \frac {7 \pi}{12} \right) \\ & = \pi \left(\cos \frac {5 \pi}{12} + i \sin \frac {5 \pi}{12} + \cos \frac {5 \pi}{12} - i\sin \frac {5 \pi}{12} \right) \\ & = 2 \pi \cos \frac {5 \pi}{12} = 2 \pi \sin \frac \pi{12} = 2 \pi \sin 15^\circ \\ & = \frac {\pi \sin 15^\circ}{\frac 12 \times 1} = \frac {\pi \sin 15^\circ}{\sin 30^\circ \sin 90^\circ} \end{aligned}$

$\implies A = \boxed{30}$

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Note:

$\begin{aligned} J_1 & = \int^\epsilon_R \frac {\sqrt z}{z^2+\sqrt 3 z + 1} dz \\ & = \int^\epsilon_R \frac {e^{\frac 12 \ln z}}{z^2+\sqrt 3 z + 1} dz \\ & = \int^\epsilon_R \frac {e^{\frac 12 (\ln |z| + i \ {\color{#3D99F6}\arg z})}}{z^2+\sqrt 3 z + 1} dz & \small \color{#3D99F6} \text{Note: } \arg z = 2\pi \text{ one revolution} \\ & = \int^\epsilon_R \frac {e^{\frac 12 \ln |z|} e^{\pi i}}{z^2+\sqrt 3 z + 1} dz \\ & = \int^\epsilon_R \frac {- e^{\frac 12 \ln |z|}}{z^2+\sqrt 3 z + 1} dz \\ & = \int^\epsilon_R \frac {- \sqrt z}{z^2+\sqrt 3 z + 1} dz \\ & = \int_\epsilon^R \frac {\sqrt z}{z^2+\sqrt 3 z + 1} dz \end{aligned}$