A calculus problem by Hamza A

$I = \displaystyle \int_{0}^{\infty} \frac{1}{1+x^6}\, dx = \displaystyle \int_{0}^{1} \frac{1}{1+x^6}\, dx + \displaystyle \int_{1}^{\infty} \frac{1}{1+x^6}\, dx$

Let $I' = \displaystyle \int_{1}^{\infty} \frac{1}{1+x^6}\, dx$

Now substituting $x = \frac{1}{p}$ and then replacing $p$ by $x$ we get:

$I' = \displaystyle \int_{0}^{1} \frac{x^4}{1+x^6}\, dx$

Therefore now we can write $I$ as :

$I=\displaystyle \int_{0}^{1} \frac{1}{1+x^6}\, dx +\displaystyle \int_{0}^{1} \frac{x^4}{1+x^6}\, dx = \displaystyle \int_{0}^{1} \frac{1+x^4}{1+x^6}\, dx$

$I = \displaystyle \int_{0}^{1} \frac{(1+x^2)^2}{1+x^6}\, dx - \displaystyle \int_{0}^{1} \frac{2x^2}{1+x^6}\, dx$

$I = \displaystyle \int_{0}^{1} \frac{1+x^2}{1-x+x^2}\, dx - \frac{2}{3} \displaystyle \int_{0}^{1} \frac{3x^2}{1+(x^3)^2}\, dx$

$I = \displaystyle \int_{0}^{1} \frac{\frac{1}{x^2} + 1}{(x-\frac{1}{x})^2 +1}\, dx - \frac{2}{3} \biggl[tan^{-1}(x^3) \biggr]_{0}^{1}$

$I = \biggl[tan^{-1}\left(x-\frac{1}{x} \right)\biggr]_0^1 - \frac{2}{3} \biggl[tan^{-1}(x^3) \biggr]_{0}^{1}$

$I = \left(0-\left(-\frac{\pi}{2}\right)\right) - \frac{2}{3}\left(\frac{\pi}{4}\right)$

$I = \frac{\pi}{3}$

Therefore $a=\boxed{3}$

$\begin{aligned} I & = \int_0^\infty \frac{1}{1+x^6} \, dx \quad \quad \quad \quad \quad \quad \small \color{#3D99F6}{\text{Let} \tan \theta = x^3 \, \Rightarrow \sec^2 \theta \, d \theta = 3x^2 \, dx} \\ & = \int_0^\frac{\pi}{2} \frac{\sec^2 \theta}{3 \tan^\frac{2}{3} \theta (1+\tan^2 \theta)} \, d\theta \\ & = \frac{1}{3} \int_0^\frac{\pi}{2} \sin^{-\frac{2}{3}} \cos^{\frac{2}{3}} \, d\theta \\ & = \frac{1}{6} \cdot{} B \left(\frac{1}{6}, \frac{5}{6} \right) \quad \quad \quad \quad \quad \quad \, \, \small \color{#3D99F6}{B(m,n) \text{ is Beta function.}} \\ & = \frac{\Gamma \left(\frac{1}{6} \right) \Gamma \left(\frac{5}{6} \right)}{6 \Gamma \left(1 \right)} \quad \quad \quad \quad \quad \quad \quad \, \small \color{#3D99F6}{\Gamma(n) \text{ is Gamma function.}} \\ & = \frac{\Gamma \left(\frac{1}{6} \right) \Gamma \left(1-\frac{1}{6} \right)}{6(\color{#3D99F6}{0!})} \quad \quad \quad \quad \quad \, \, \small \color{#3D99F6}{\Gamma(n) = (n-1)!} \\ & = \frac{1}{6} \cdot{} \color{#3D99F6}{\frac{\pi}{\sin \left(\frac{\pi}{6} \right)}} \quad \quad \quad \quad \quad \quad \quad \, \, \small \color{#3D99F6}{\Gamma(x) \Gamma (1-x) = \frac{\pi}{\sin \left(x \pi \right)}} \\ & = \frac{\pi}{3} \end{aligned}$

$\Rightarrow a = \boxed{3}$

Using the Residue Theorem which states $\displaystyle\oint\limits_\kappa f(z)dz = 2\pi i \sum_{}^{}Res(f(z),z_{0})$

$\displaystyle\oint\limits_\kappa f(z)dz = \displaystyle\int_{-R}^{R} f(z)dz + \displaystyle\int\limits_\kappa f(z)dz$

$\displaystyle\lim_{R\to\infty}\quad\int\limits_\kappa f(z)dz = 0$

Using a semicircle contour, $\kappa$ , in the $1^{st}$ and $2^{nd}$ quadrants of radius $R$ allows only the usage of the poles in this region which are $e^{\frac{\pi i}{6}},e^{\frac{\pi i}{2}},$ and $e^{\frac{5\pi i}{6}}$ .

$\alpha = e^{\frac{\pi i}{6}}$

The residue of $f(z)=\frac{1}{z^6+1}$ at a pole is, $\displaystyle\lim_{z\to\beta}\quad (z-\beta) f(z)$ , where $\beta$ is a pole, to find the $a_{-1}$ term in the Laurent series.

$\lim_{z\to\alpha}\quad \frac{z-\alpha}{z^6+1} = \frac{e^{\frac{11\pi i}{6}}}{6}$ and repeating for $\alpha^3$ and $\alpha^5$ produces the sum of the residues in $\kappa$ :

$\frac{e^{\frac{11\pi i}{6}}}{6}+\frac{e^{\frac{3\pi i}{2}}}{6}+\frac{e^{\frac{5\pi i}{6}}}{6} = \frac{-i}{6}$ , but there are 3 more poles on another semi-circle contour on the $3^{rd}$ and $4^{th}$ quadrants. They are obtainable by rotating $\kappa$ by $180˚$ but multiplying the sum by 2 would do. Therefore the sum of all residues is $\frac{-i}{3}$

$\displaystyle 2\pi i \times \frac{-i}{3} = \frac{2\pi}{3}$

The function is even so $\displaystyle\int_{-a}^{a}f(x)dx = 2\int_{0}^{a}f(x)dx$ .

$\displaystyle\int_{0}^{\infty}f(x)dx = \boxed{\frac{\pi}{3}}$