How to find the exact value of the integral $ \int_{0}^{\infty} \frac{d x}{\left(x^{3}+\frac{1}{x^{3}}\right)^{2}}$?

Question

$\textrm{I first reduce the power two to one by Integration by Parts.}$

$\begin{aligned}\displaystyle \int_{0}^{\infty} \frac{1}{\left(x^{3}+\frac{1}{x^{3}}\right)^{2}} d x &=\int_{0}^{\infty} \frac{x^{6}}{\left(x^{6}+1\right)^{2}} d x\\&=\displaystyle -\frac{1}{6} \int_{0}^{\infty} x d\left(\frac{1}{x^{6}+1}\right)\\& =\displaystyle -\left[\frac{x}{6\left(x^{6}+1\right)}\right]_{0}^{\infty}+\frac{1}{6} \int_{0}^{\infty} \frac{1}{x^{6}+1} d x \quad \textrm{ (Via Integration by Parts})\\&=\displaystyle \frac{1}{6} \int_{0}^{\infty} \frac{1}{x^{6}+1} d x\end{aligned}$

$\textrm{Then I am planning to evaluate }\displaystyle I= \int_{0}^{\infty} \frac{1}{x^{6}+1}\text{ by resolving }\frac{1}{x^{6}+1} \text{ into partial fractions.}$

But after noticing that $$I=\int_{0}^{\infty} \frac{d x}{x^{6}+1}\stackrel{x\mapsto\frac{1}{x}}{=} \int_{0}^{\infty} \frac{x^{4}}{x^{6}+1} d x,$$

I changed my mind and started with $3I$ instead of $I$ as below:

$$ \begin{aligned} 3 I &=\int_{0}^{\infty} \frac{x^{4}+2}{\left(x^{2}+1\right)\left(x^{4}-x^{2}+1\right)} d x \\ &=\int_{0}^{\infty}\left(\frac{1}{x^{2}+1}+\frac{1}{x^{4}-x^{2}+1}\right) d x \\ &=\left[\tan ^{-1} x\right]_{0}^{\infty}+\int_{0}^{\infty} \frac{\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}-1} d x \\ &=\frac{\pi}{2}+\frac{1}{2} \int_{0}^{\infty} \frac{\left(1+\frac{1}{x^{2}}\right)-\left(1-\frac{1}{x^{2}}\right)}{x^{2}+\frac{1}{x^{2}}-1} d x\\ &=\frac{\pi}{2}+\frac{1}{2}\left[\int_{0}^{\infty} \frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^{2}+1}-\int \frac{d\left(x+\frac{1}{x}\right)}{\left(x+\frac{1}{x}\right)^{2}-3}\right] \\ &=\frac{\pi}{2}+\frac{1}{2}\left[\tan ^{-1}\left(x-\frac{1}{x}\right)\right]_{0}^{\infty}-0 \\ &=\frac{\pi}{2}+\frac{1}{2}\left[\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)\right] \\ &=\pi \\ \therefore I &=\frac{\pi}{3} \end{aligned} $$ Now I can conclude that $$\boxed{\displaystyle \quad \int_{0}^{\infty} \frac{1}{\left(x^{3}+\frac{1}{x^{3}}\right)^{2}} d x=\frac{\pi}{18} }.$$

:|D Wish you enjoy the solution! Opinions and alternative methods are welcome.

Answer 1

An alternative method using some known identities .

$\begin{aligned}\displaystyle \int_{0}^{\infty} \frac{1}{\left(x^{3}+\frac{1}{x^{3}}\right)^{2}} d x &=\int_{0}^{\infty} \frac{x^{6}}{\left(x^{6}+1\right)^{2}} d x.\end{aligned}$

Let $\displaystyle x=t^{\frac{1}{6}}$.

Then we get $$\int_{0}^{\infty}\frac{1}{6}\frac{t\cdot t^{\frac{-5}{6}}}{(1+t)^{2}}dt=\int_{0}^{\infty}\frac{1}{6}\frac{t^{\frac{1}{6}}}{(1+t)^{2}}dt.$$

We know that $\text{B}(m,n)=\int_{0}^{\infty}\frac{x^{m-1}}{(1+x)^{m+n}}dx.$

So we see that our integral is nothing but: $$\frac{1}{6}\text{B}(\frac{7}{6},\frac{5}{6}).$$

Now expanding we get

$$\frac{1}{6}\frac{\Gamma(\frac{7}{6})\Gamma(\frac{5}{6})}{\Gamma(2)}=\frac{1}{6}\frac{\frac{1}{6}\Gamma(\frac{1}{6})\Gamma(\frac{5}{6})}{1}.$$

Now using Euler's reflection formula. We get: $\Gamma(n)\Gamma(1-n)=\pi\csc(n\pi)$

$$\frac{1}{36}\csc(\frac{\pi}{6})=\frac{\pi}{18}.$$

This is just an alternate method. It's no way even close to being as elegant as your solution. But it is perhaps a little easier for people who know these identities.

Answer 2

First, evaluate \begin{align}\int_{0}^{\infty} \frac{1}{x^{6}+1}dx\stackrel{x\to\frac{1}{x}}{=}&\ \frac12 \int_{0}^{\infty} \frac{x^{4}+1}{x^{6}+1} d x\\ =&\ \frac12 \bigg(\int_{0}^{\infty} \frac{1}{x^{2}+1} d x + \int_{0}^{\infty} \frac{x^2}{x^{6}+1}\overset{x^3\to x}{ d x}\bigg)\\ =&\ \frac12\cdot \frac43\int_{0}^{\infty} \frac{1}{x^2+1}dx =\frac\pi3 \end{align} Then, \begin{aligned}\int_{0}^{\infty} \frac{1}{\left(x^{3}+\frac{1}{x^{3}}\right)^{2}} d x \overset{ibp}=\frac{1}{6} \int_{0}^{\infty} \frac{1}{x^{6}+1} d x =\frac\pi{18}\end{aligned}

Answer 3

It is a nice solution for sure.

As you showed the problem is to compute $$I=\int \frac {dx}{x^6+1}$$

My working

Writing $$x^6+1=(x^2+1)(x^4-x^2+1)=(x^2+1)(x^2-a)(x^2-b)$$ Using partial fraction decomposition $$\frac 1 {(x^2+1)(x^2-a)(x^2-b)}=$$ $$\frac{1}{(a+1) (a-b) \left(x^2-a\right)}-\frac{1}{(b+1) (a-b) \left(x^2-b\right)}+\frac{1}{(a+1) (b+1) \left(x^2+1\right)}$$ Integrating $$I=\frac{\tan ^{-1}(x)}{(a+1) (b+1)}-\frac 1{a-b}\Bigg[\frac{\tanh ^{-1}\left(\frac{x}{\sqrt{a}}\right)}{(a+1)\sqrt{a} }-\frac{\tanh ^{-1}\left(\frac{x}{\sqrt{b}}\right)}{(b+1)\sqrt{b} } \Bigg]$$ Using the bounds, we then have $$J=\int_0^\infty \frac {dx}{x^6+1}=\frac{\pi }{2 (a+1) (b+1)}+\frac \pi{2(a-b)}\Bigg[\frac{\sqrt{-\frac{1}{a}}}{a+1}-\frac{\sqrt{-\frac{1}{b}}}{b+1}\Bigg]$$ $$a=\frac{1+i\sqrt{3}}{2}= \sqrt{-\frac{1}{a}}\qquad \text{and} \qquad b=\frac{1-i\sqrt{3}}{2}= \sqrt{-\frac{1}{b}} $$ So, finally $$J=\frac{\pi }{1+(a+b)+a b}=\frac \pi 3$$

Answer 4

In general

$$ \int_{0}^{\infty} \frac{1}{1+x^s} dx = \frac{\pi}{s}\csc\left(\frac{\pi}{s}\right) \quad \Re(s)>1$$

$$\int_{0}^{\infty} \frac{1}{1+x^s}dx = \frac{2}{s}\int_{0}^{\infty} \frac{w^{\frac{2}{s}-1}}{1+w^2}dw \quad (w^2 \mapsto x^s)$$

Recall the integral representation of $\displaystyle \sec z$:

$$ \sec z = \frac{2}{\pi}\int_{0}^{\infty} \frac{t^{\frac{2z}{\pi}}}{1+t^2} dt \quad |\Re(z)|<\frac{\pi}{2}$$

Putting $ \displaystyle z = \frac{\pi}{s} -\frac{\pi}{2}$

$$ \frac{\pi}{2} \sec\left(\frac{\pi}{s} -\frac{\pi}{2} \right) = \frac{\pi}{2} \csc\left(\frac{\pi}{s}\right)= \int_{0}^{\infty} \frac{t^{\frac{2}{s}-1}}{1+t^2} dt $$

$$ \Longrightarrow \frac{\pi}{s}\csc\left(\frac{\pi}{s}\right) = \frac{2}{s} \int_{0}^{\infty} \frac{t^{\frac{2}{s}-1}}{1+t^2} dt \quad \Re(s)> 1 $$

Hence

$$\boxed{ \int_{0}^{\infty} \frac{1}{1+x^s}dx =\frac{\pi}{s}\csc\left(\frac{\pi}{s}\right) \quad \Re(s)> 1 }$$

Therefore

$$\boxed{ \int_{0}^{\infty} \frac{1}{1+x^6}dx =\frac{\pi}{6}\csc\left(\frac{\pi}{6}\right) = \frac{\pi}{3} }$$