Solving $\int\limits_{-\infty}^\infty \frac{1}{x^8+1}dx$ through Glasser's Master Theorem

Question

Trying to find a way to solve

$$\int_{-\infty}^\infty \frac{1}{x^8+1}dx$$

through Glasser's Master Theorem, more specifically the Cauchy–Schlömilch substitution. Preferably, I'm looking for the closed form solution, and I am already aware of how to attain this through contour integration.

Solution: $$\frac{\pi}{4\sin(\frac{\pi}{8})}$$

Link to general closed form solution: solutions to $\int_{-\infty}^\infty \frac{1}{x^n+1}dx$ for even $n$

Answer 1

I don't know what is GMT. I looked it up on Wikipedia and it states $u = x - 1/x$ as Cauchy–Schlömilch substitution. Here is my solution that splits the integral into 4 integrals, each one of them uses $u = x - 1/x$. I will happily delete the answer if it is useless or wrong :)

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$$\int^{\infty}_{-\infty} \dfrac{1}{1+ x^8} dx = -\dfrac{1}{2 \sqrt{2}}\int^{\infty}_{-\infty} \dfrac{x^2 - \sqrt{2}}{x^4 - \sqrt 2 x^2 + 1} + \dfrac{1}{2 \sqrt{2}}\int^{\infty}_{-\infty}\dfrac{x^2 + \sqrt{2}}{x^4 + \sqrt2 x^2 + 1} \\= -\dfrac{1}{2 \sqrt{2}}\int^{\infty}_{-\infty} \dfrac{x^2 + 1}{x^4 - \sqrt 2 x^2 + 1} +\dfrac{\sqrt{2} +1}{2 \sqrt{2}} \int^{\infty}_{-\infty} \dfrac{1}{x^4 - \sqrt 2 x^2 + 1} \\+ \dfrac{1}{2 \sqrt{2}}\int^{\infty}_{-\infty}\dfrac{x^2 + 1 }{x^4 + \sqrt2 x^2 + 1}+\dfrac{\sqrt2 -1}{2 \sqrt{2}}\int^{\infty}_{-\infty}\dfrac{1}{x^4 + \sqrt2 x^2 + 1} $$

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$$\int^{\infty}_{-\infty} \dfrac{x^2 + 1}{x^4 - \sqrt 2 x^2 + 1} = \int^{\infty}_{-\infty} \dfrac{1 + 1/x^2}{ (x -1/x)^2 + 2 - \sqrt 2}$$

$u = x - 1/x, du = 1 + 1/x^2$

$$\int^{\infty}_{-\infty} \dfrac{1}{u^2 + (2 - \sqrt 2 )} du $$

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$$J = \int^{\infty}_{-\infty} \dfrac{1}{x^4 - \sqrt 2 x^2 + 1} = 2 \int^{\infty}_{0} \dfrac{1}{x^4 - \sqrt 2 x^2 + 1}$$

$x = 1/t, dx = -1/t^2 dt$

$$J = 2 \int^{\infty}_{0} \dfrac{t^2}{t^4 - \sqrt 2 t^2 + 1} dt$$

Adding the two equations, $$J = \int^{\infty}_{0} \dfrac{t^2 + 1}{t^4 - \sqrt 2 t^2 + 1} dt$$

This can be solved with $ u = t - 1/t$ like the previous equation.

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$$\int^{\infty}_{-\infty}\dfrac{x^2 + 1 }{x^4 + \sqrt2 x^2 + 1} = \int^{\infty}_{-\infty}\dfrac{1 + 1/x^2 }{(x - 1/x)^2 + \sqrt2 + 2}$$

Similar to first integral, now.

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$$I = \int^{\infty}_{-\infty}\dfrac{1}{x^4 + \sqrt2 x^2 + 1} = 2\int^{\infty}_{0}\dfrac{1}{x^4 + \sqrt2 x^2 + 1}$$

$x = 1/t, dx = -1/t^2 dt$

$$ I = \int^{\infty}_{0}\dfrac{t^2 + 1}{t^4 + \sqrt2 t^2 + 1} dt $$

Can be solved with $u = t - 1/t$ like other integrals here.

Answer 2

I believe the most direct way for tackling such integrals is to exploit Euler's Beta function and the reflection formula for the $\Gamma$ function: assuming $m>1$,

$$ \int_{0}^{+\infty}\frac{dx}{1+x^m}\stackrel{\frac{1}{1+x^m}\to u}{=}\frac{1}{m}\int_{0}^{1}u^{-\frac{m-1}{m}}(1-u)^{-\frac{1}{m}}\,du =\frac{\Gamma\left(\tfrac{1}{m}\right)\Gamma\left(\tfrac{m-1}{m}\right)}{m\,\Gamma(1)}=\frac{\pi/m}{\sin(\pi/m)}.$$ Another way is to exploit the reflection formula for the digamma function, in the form $$ \sum_{n\geq 0}\left[\frac{1}{an+b}+\frac{1}{an+(a-b)}\right]=\frac{\pi}{a}\,\cot\left(\frac{\pi b}{a}\right)$$ via $$\begin{eqnarray*} \int_{0}^{+\infty}\frac{dx}{1+x^m}&=&\int_{0}^{1}\frac{1+x^{m-2}}{1+x^m}\,dx=\int_{0}^{1}\frac{1+x^{m-2}-x^m-x^{2m-2}}{1-x^{2m}}\,dx\\ &=& \sum_{n\geq 0}\left[\color{blue}{\frac{1}{2mn+1}}+\color{red}{\frac{1}{2mn+m-1}}-\color{red}{\frac{1}{2mn+m+1}}-\color{blue}{\frac{1}{2mn+2m-1}}\right]\\&=&\frac{\pi}{2m}\left[\cot\left(\frac{\pi}{2m}\right)+\tan\left(\frac{\pi}{2m}\right)\right].\end{eqnarray*}$$ Through Glasser's master theorem, in the $m=8$ case we may state

$$\begin{eqnarray*} \int_{-\infty}^{+\infty}\frac{dx}{1+x^8}&\stackrel{x\mapsto z\cdot 2^{1/4}}{=}&\int_{-\infty}^{+\infty}\frac{dz}{(1-\sqrt{2}z^2+z^4)(1+\sqrt{2}z^2+z^4)} \\&=&2\int_{0}^{+\infty}\frac{1}{2\sqrt{2}z^2}\left[\frac{1}{1-\sqrt{2}z^2+z^4}-\frac{1}{1+\sqrt{2}z^2+z^4}\right]\,dz \\&\stackrel{z\mapsto 1/z}{=}&2\int_{0}^{+\infty}\frac{z^4}{2\sqrt{2}}\left[\frac{1}{1-\sqrt{2}z^2+z^4}-\frac{1}{1+\sqrt{2}z^2+z^4}\right]\,dz \\&=&\frac{1}{\sqrt{2}}\int_{0}^{+\infty}\frac{1}{z^4}\left[\frac{1}{\frac{1}{z^2}-\sqrt{2}+z^2}-\frac{1}{\frac{1}{z^2}+\sqrt{2}+z^2}\right]\,dz \\&=&\frac{1}{\sqrt{2}}\int_{0}^{+\infty}z^2\left[\frac{1}{\frac{1}{z^2}-\sqrt{2}+z^2}-\frac{1}{\frac{1}{z^2}+\sqrt{2}+z^2}\right]\,dz\end{eqnarray*}$$ then invoke integration by parts and averaging in order to convert the original integral into $$\begin{eqnarray*}\int_{0}^{+\infty}r\left(x^2+\frac{1}{x^2}\right)\,dx &=& \frac{1}{2}\int_{-\infty}^{+\infty}r\left(\left(x-\frac{1}{x}\right)^2+2\right)\,dx\\&\stackrel{\text{GMT}}{=}&\frac{1}{2}\int_{-\infty}^{+\infty}r(x^2+2)\,dx=\int_{0}^{+\infty}r(x^2+2)\,dx\end{eqnarray*}$$ with $r$ being a rational function. On the other hand, this approach looks pretty forced/artificial, especially if compared to the previous ones.

Answer 3

I prefer to use Ramanujan's Master Theorem in situations like this which is a specific Master Theorem for the Mellin Transform.

Ramanujan's Master Theorem

Let $f(x)$ be an analytic function with a MacLaurin Expansion of the form $$f(x)=\sum_{k=0}^{\infty}\frac{\phi(k)}{k!}(-x)^k$$then the Mellin Transform of this function is given by $$\int_0^{\infty}x^{s-1}f(x)dx=\Gamma(s)\phi(-s)$$

The structure of the integrande remindes us the geometric series. First of all noting that we have an even function followed by the substitution $u=x^8$ yields to

$$\begin{align*} \mathfrak I=\int_{-\infty}^\infty \frac{\mathrm dx}{1+x^8}&=2\int_0^\infty \frac{\mathrm dx}{1+x^8}\\ &=2\int_0^\infty \sum_{n=0}^\infty (-x^8)^{n}\mathrm dx\\ &=2\int_0^\infty \sum_{n=0}^\infty (-u)^{n}\frac{\mathrm du}{8u^{7/8}}\\ &=\frac14 \int_0^\infty u^{-7/8}\sum_{n=0}^\infty \frac{n!}{n!}(-u)^{n}\mathrm du \end{align*}$$

Now we can use Ramanujan's Master Theorem with $s=\frac18$ and $\phi(n)=\Gamma(n+1)$ to get

$$\begin{align*} \mathfrak I=\frac14 \int_0^\infty u^{1/8-1}\sum_{n=0}^\infty \frac{\Gamma(n+1)}{n!}(-u)^{n}\mathrm du&=\frac14\Gamma\left(\frac18\right)\Gamma\left(1-\frac18\right)\\ &=\frac14\frac\pi{\sin\left(\frac\pi8\right)} \end{align*}$$

$$\therefore~\mathfrak I=\int_{-\infty}^\infty \frac{\mathrm dx}{1+x^8}~=~\frac\pi{4\sin\left(\frac\pi8\right)}$$

The latter manipulations are done via Euler's Reflection Formula. Overall this approach is pretty straightforward and can be easily generalised for arbitrary powers of $x$.

Answer 4

NOT A SOLUTION:

In case you are interested in another 'real' based method to solve this - I actually posted a question on this matter (with my solution) yesterday.

In terms of Glasser's Master Theorem (GMT), I'm not sure it can be positioned that way to be honest... or better put that it would require some very careful (and algebraically extensive) work to yield it into the desired form. In saying that, I could be very wrong about that.

Have you considered starting with a form that is GMT compliant (with free parameters) and attempting to solve? I tried with a few forms and was unable to yield a solution.

In terms of compliant forms, I tried to 'reverse engineer' the result, so I started with the following expressions and tried to solve for the unknown constants:

\begin{equation} \frac{1}{x^8 + 1} = \left[\left(x - \frac{b_1}{x - c_1}\right)^4 + d_1 \right]^{-1} \end{equation} This didn't work, so I tried: \begin{equation} \frac{1}{x^2 + 1} = \left[\left(x - \frac{b_1}{x - c_1} - \frac{b_2}{x - c_2}\right)^4 + d_1 \right]^{-1} \end{equation}

I tried a few others, but this was the method employed. As before, it was unsuccessful for me and so I tried different ways. Shame as it seems so close to being GMT applicable!

Answer 5

Decompose the integrand to express the integral as

\begin{align} I=&\int_{-\infty}^\infty \frac1{1+x^8}dx \\ =& \frac1{2\sqrt2}\int_{-\infty}^\infty \frac{\sqrt2+x^2}{x^4+\sqrt2 x^2+1}dx + \frac1{2\sqrt2}\int_{-\infty}^\infty \frac{\sqrt2-x^2}{x^4+\sqrt2 x^2+1}dx\\ \overset{x\to \frac1x}=& \frac1{2\sqrt2}\int_{-\infty}^\infty \frac{(\sqrt2+1)x^2}{x^4+\sqrt2 x^2+1}dx + \frac1{2\sqrt2}\int_{-\infty}^\infty \frac{(\sqrt2-1)x^2}{x^4+\sqrt2 x^2+1}dx\\ =& \frac{\sqrt2+1}{2\sqrt2}\int_{-\infty}^\infty \frac{1}{(x-\frac1x)^2+2+\sqrt2}dx + \frac{\sqrt2-1}{2\sqrt2}\int_{-\infty}^\infty \frac{1}{(x-\frac1x)^2+2-\sqrt2}dx\\ \end{align} Then, apply the Glasser master theorem $\int_{-\infty}^\infty f(x-\frac1x)dx= \int_{-\infty}^\infty f(x)dx$ \begin{align} I =& \frac{\sqrt2+1}{2\sqrt2}\int_{-\infty}^\infty \frac1{x^2+(2+\sqrt2)}dx + \frac{\sqrt2-1}{2\sqrt2}\int_{-\infty}^\infty \frac{1}{x^2+(2-\sqrt2)}dx\\ =& \frac{\sqrt2+1}{2\sqrt2}\cdot \frac{\pi}{\sqrt{2+\sqrt2}} + \frac{\sqrt2-1}{2\sqrt2}\cdot \frac{\pi}{\sqrt{2-\sqrt2}} =\frac\pi2 \sqrt{1+\frac1{\sqrt2}}=\frac\pi4 \csc\frac\pi8 \end{align}