Calculate :$\int_{0}^{\infty}{ }\frac{dx}{a^6+(x-\frac{1}{x})^6}$ Find the:
$$\int_{0}^{\infty}{ }\frac{dx}{a^6+(x-\frac{1}{x})^6}:a>0$$
My Try:
$$\frac{1}{a^6}\int_{0}^{\infty}{ }\frac{dx}{1+(\frac{x-\frac{1}{x}}{a})^6}$$
$$\int_0^\infty{dx\over1+u^6}={1\over2}\left({\pi\over2}+{\pi\over6}+0\right){\pi\over3}$$
Is it right?
 A: By Glasser's master theorem the given integral equals
$$ \int_{0}^{+\infty}\frac{dx}{a^6+x^6} = \frac{1}{|a|^5}\int_{0}^{+\infty}\frac{du}{1+u^6} $$
that by Euler's Beta function and the $\Gamma$ reflection formula equals $\color{red}{\large\frac{\pi}{3|a|^5}}$.
A: Let $$I = \int^{\infty}_{0}\frac{1}{a^6+\bigg(x-\frac{1}{x}\bigg)^6}dx$$
substitute $\displaystyle x= \frac{1}{t}$ Then $$I = \int^{\infty}_{0}\frac{1}{a^6+\bigg(t-\frac{1}{t}\bigg)^6}\cdot \frac{1}{t^2}dt = \int^{\infty}_{0}\frac{1}{a^6+\bigg(x-\frac{1}{x}\bigg)^6}\cdot \frac{1}{x^2}dx$$
So $$2I = \int^{\infty}_{0}\frac{1}{a^6+\bigg(x-\frac{1}{x}\bigg)^6}\cdot \bigg(1+\frac{1}{x^2}\bigg)dx$$
Substitute $\displaystyle \bigg(x-\frac{1}{x}\bigg)=u,$ then $\displaystyle \bigg(1+\frac{1}{x^2}\bigg)dx = du$
so $$2I = \int^{\infty}_{-\infty}\frac{1}{a^6+u^6}du = 2\int^{\infty}_{0}\frac{1}{a^6+u^6}du=\frac{2}{a^6}\int^{\infty}_{0}\frac{1}{1+\left(\frac{u}{a}\right)^6}du$$
put  $\displaystyle \frac{u}{a} = v\;,$ then $\displaystyle I = \frac{1}{a^5}\int^{\infty}_{0}\frac{1}{1+v^6}dv = \frac{\pi}{6a^5}\cdot \frac{1}{\sin \frac{\pi}{6}} = \frac{2\pi}{6a^5} = \frac{\pi}{3a^5}$
Show that $\int_0^ \infty \frac{1}{1+x^n} dx= \frac{ \pi /n}{\sin(\pi /n)}$ , where $n$ is a positive integer.
A: By Glasser's master theorem, the integral reduces to
$$\int_0^{+\infty}\frac1{a^6+x^6}\ dx=\frac1{|a|^5}\int_0^{+\infty}\frac1{1+u^6}\ du$$
Which, by taking a semicircle contour in the upper half of the complex plane, we end up with:
$$\int_0^{+\infty}\frac1{1+x^6}\ du=-\frac{\pi i}6(e^{\pi i/6}+e^{3\pi i/6}+e^{5\pi i/6})=\frac\pi3$$
Thus,
$$\int_0^{+\infty}\frac1{a^6+\left(x-\frac1x\right)^6}\ dx=\frac\pi{3|a|^5}$$
