Question:

Find the value of integral

\[ \int^{\infty}_{-\infty}\frac{dx}{x^6+1} \]

Answer:

The integrand is even function, so we can reduce the above integral into

\[ \int^{\infty}_{-\infty}\frac{dx}{x^6+1} = 2\int^{\infty}_{0}\frac{dx}{x^6+1} \]

Then, replace \(x\) with \(t^{\frac{1}{6}}\). Thus, we have \(dx=\frac{1}{6}t^{-\frac{5}{6}}dt\). Thus,

\[ \int^{\infty}_{0}\frac{dx}{x^6+1} = \frac{1}{6}\int^{\infty}_{0}\frac{t^{-\frac{5}{6}}}{t+1}dt \]

Remember that the Beta function is defined as follows:

\[ B(p,q) = \int^{\infty}_{0}\frac{x^{p-1}}{(1+x)^{p+q}}dx \]

\[ B(p,q)=B(q,p)=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)} \]

So the above expression gives you

\[ \frac{1}{6}\int^{\infty}_{0}\frac{t^{-\frac{5}{6}}}{t+1}dt = \frac{1}{6}B(\frac{1}{6},\frac{5}{6}) \]

\[   = \frac{1}{6}\frac{\Gamma(\frac{1}{6})\Gamma(\frac{5}{6})}{\Gamma(\frac{1}{6}+\frac{5}{6})} \]

\[  =  \frac{1}{6}\Gamma(\frac{1}{6})\Gamma(\frac{5}{6})  \]

where we use \(\Gamma(1)=1\).

Do you remember the following formula?

\[ \Gamma(s)\Gamma(1-s) = \frac{\pi}{\sin\pi s}  \]

The parameter satisfies \(0<s<1\).

Thus,

\[ \frac{1}{6}\Gamma(\frac{1}{6})\Gamma(\frac{5}{6}) = \frac{1}{6}\frac{\pi}{\sin(\pi/6)} = \frac{\pi}{3} \]

Hence,

\[ \int^{\infty}_{-\infty}\frac{dx}{x^6+1} = \frac{2\pi}{3}\]