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 $n$

  • $\begingroup$ you mean the indefinite integral ? Because the closed form is just a number otherwise .. $\endgroup$ – Malik Dec 19 '18 at 0:24
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    $\begingroup$ @Malik I mean closed form as in "pi/e" or something an not just a decimal $\endgroup$ – Suchetan Dontha Dec 19 '18 at 0:30
  • $\begingroup$ (Seeing as you do know the answer, you might as well include it in the question.) $\endgroup$ – Ian Dec 19 '18 at 0:30
  • $\begingroup$ @Ian okay sure thing $\endgroup$ – Suchetan Dontha Dec 19 '18 at 0:31
  • $\begingroup$ Is it definitely the case that there is a route through Cauchy–Schlömilch? $\endgroup$ – Mason Dec 19 '18 at 0:44

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 :)

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

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

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

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

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

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

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

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.

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

Similar to first integral, now.

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

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

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

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

  • $\begingroup$ Thanks for your solution, although the Cauchy–Schlömilch substitution states that $\int_{-\infty}^\infty F(u)dx = \int_{-\infty}^\infty F(x)dx$, notice the "F(u)dx". This is not regular u substitution, but still plus +1 for the effort. Please don't delete it. $\endgroup$ – Suchetan Dontha Dec 19 '18 at 13:59
  • $\begingroup$ I won't, no worries :) $\endgroup$ – user8277998 Dec 19 '18 at 15:12

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.



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!

  • $\begingroup$ Thanks for the info! And you may very well be right, it may not be possible in the form it is right now. What do you mean by starting in a form that is GMT compliant? $\endgroup$ – Suchetan Dontha Dec 19 '18 at 4:00
  • $\begingroup$ @SuchetanDontha I've been faced with this before. I desperately tried to get it to work but couldn't (not to say it can't be done). Re Compliant - I will add more detail in my 'answer'. Am just in a meeting right now, but will be edit in 15 minutes. $\endgroup$ – user150203 Dec 19 '18 at 4:07
  • $\begingroup$ @SuchetanDontha - I have added more detail :-) $\endgroup$ – user150203 Dec 19 '18 at 4:27
  • $\begingroup$ Thanks for the tips, although I don't have that much math experience under my belt, so brute force is the only way I can go! I think the difficult part here is the sin(pi/8) on the bottom of the solution. Even for x^6, the closed form solution simplifies to a nice pi times a coefficient, so i think some sort of trigonometric argument would need to come into play. However, I still believe it is possible and will keep trying. $\endgroup$ – Suchetan Dontha Dec 19 '18 at 14:10
  • $\begingroup$ Excellent. When you get it out please post and tag me. $\endgroup$ – user150203 Dec 19 '18 at 14:17

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 $x$.

  • $\begingroup$ How is $\phi(s)$ defined when $s$ is not a nonnegative integer? // Later on, the substitution $w=\frac1{1+x^8}$ works more quickly and simply. $\endgroup$ – Did Jan 8 at 8:24
  • $\begingroup$ @Did I am not exactly sure concerning your question: are you asking for the general case or regarding to this particular application? Hence for the first situation one have to consider that the integral representation $-$ as Ramanujan used it in the first place $-$ only works out for $0<\Re(s)<1$. $\endgroup$ – mrtaurho Jan 8 at 8:52
  • $\begingroup$ There are tons of different functions $\phi$ such that the first identity in your paragraph "Ramanujan's Master Theorem" holds, but not all of them can make its conclusion true when $s$ is not an integer. For example, if $\phi$ works, then $\phi_1:s\mapsto\phi(s)+7\sin(\pi s)$ works as well. So, which one should one choose for the conclusion to hold? $\endgroup$ – Did Jan 8 at 8:56
  • $\begingroup$ @Did To be honest I am a little bit confused right now. What exactly is the aim of your commentary? Could you maybe rephrase your last question? $\endgroup$ – mrtaurho Jan 8 at 9:01
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    $\begingroup$ Yes it is, or rather, you do not understand it. Actually, you seem to be merely reproducing some wrong formulations from the WP page, neglecting the fact that, in RMT, $\phi$ is not just any function such that the expansion of $f$ holds. $\endgroup$ – Did Jan 8 at 9:35

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