Evaluation of $\int_{0}^{\infty}\frac{1}{\sqrt{x^4+x^3+x^2+x+1}}dx$ Evaluation of Integral $\displaystyle \int_{0}^{\infty}\frac{1}{\sqrt{x^4+x^3+x^2+x+1}}dx$
$\bf{My\; Try::}$ First we will convert $x^4+x^3+x^2+x+1$ into closed form, which is $\displaystyle \left(\frac{x^5-1}{x-1}\right)$
So Integral is $\displaystyle \int_{0}^{\infty}\frac{\sqrt{x-1}}{\sqrt{x^5-1}}dx$
now i did not understand how can i solve it
Help me
Thanks
 A: We have:
$$I=\int_{0}^{+\infty}\frac{dx}{\sqrt{1+x+x^2+x^3+x^4}}=2\int_{0}^{1}\frac{dx}{\sqrt{1+x+x^2+x^3+x^4}}$$
(just split $[0,+\infty)=[0,1)\cup[1,+\infty)$ and use the substitution $x=1/y$ on the second piece) and since:
$$ \sqrt{1-x}=\sum_{j=0}^{+\infty}\binom{1/2}{j}(-1)^j x^j,\qquad\frac{1}{\sqrt{1-x^5}}=\sum_{j=0}^{+\infty}\binom{-1/2}{j}(-1)^j x^{5j}$$
we can compute the integral by considering the Cauchy product of the last two series and integrating it termwise:
$$ \sqrt{\frac{1-x}{1-x^5}}=\sum_{j=0}^{+\infty}\sum_{k=0}^{\lfloor j/5\rfloor}\binom{-1/2}{k}\binom{1/2}{j-5k}(-1)^j x^j,$$
$$ I = 2\sum_{j=0}^{+\infty}\sum_{k=0}^{\lfloor j/5\rfloor}\binom{-1/2}{k}\binom{1/2}{j-5k}\frac{(-1)^j}{j+1}.$$
As an alternative, since:
$$\int_{0}^{1}x^{5j}(1-x)^{1/2}\,dx = B(5j+1,3/2) = \frac{\sqrt{\pi}\,\Gamma(5j+1)}{2\,\Gamma(5j+5/2)}$$
it follows that:

$$ I = 2\sum_{j=0}^{+\infty}\binom{-1/2}{j}(-1)^j\frac{\sqrt{\pi}\,\Gamma(5j+1)}{2\,\Gamma(5j+5/2)}=\sum_{j=0}^{+\infty}\frac{\Gamma(j+1/2)\Gamma(5j+1)}{\Gamma(j+1)\Gamma(5j+5/2)}.$$

A: $$
I=\int^{\infty}_{0}\frac{dx}{\sqrt{x^4+x^3+x^2+x+1}}
$$
Making the change of variable $x\rightarrow y/(1+y)$ we get
$$
I=\int^{1}_{0}\frac{dy}{\sqrt{y^4-2y^3+4y^2-3y+1}}=
$$
$$
=\int^{1}_{0}\frac{dy}{\sqrt{1/16(2y-1)^4+5/8(2y-1)^2+5/16}}.
$$
Also setting $(2y-1)\rightarrow u$, then
$$
I=\frac{1}{2}\int_{-1}^{1}\frac{du}{\sqrt{1/16 (u^2+5)^2-5/4}}=\int_{0}^{1}\frac{du}{\sqrt{1/16 (u^2+5)^2-5/4}}
$$
Then setting $(u^2+5)/4\rightarrow w$, we get
$$
I=\int_{5/4}^{6/4}\frac{dw}{\sqrt{w^2-5/4}}\frac{1}{\sqrt{w-5/4}}=\int_{0}^{1/4}\frac{w^{-1/2}dw}{\sqrt{(w+5/4-\sqrt{5/4})(w+5/4+\sqrt{5/4})}}=
$$
$$
=-2i\frac{5-\sqrt{20}}{\sqrt{5}\sqrt{5-\sqrt{20}}}\int^{-1/(5-\sqrt{20})}_{0}\frac{t^{-1/2}}{\sqrt{(1-t)(1-mt)}}dt,
$$
where $m=\frac{5-\sqrt{20}}{5+\sqrt{20}}$. From the deffinition of the elliptic integrals (in Mathematica's notation):
$$
F(\sin\phi,m)=\int_{0}^{\phi}\frac{1}{\sqrt{(1-t^2)(1-mt^2)}}dt=\frac{1}{2}\int_{0}^{ \phi^2}\frac{1}{\sqrt{t(1-t)(1-mt)}}dt\Leftrightarrow
$$
$$
2F\left(\sin\sqrt{\phi},m\right)=\int^{ \phi}_{0}\frac{1}{\sqrt{t(1-t)(1-mt)}}dt.
$$
Hence from (1) and (2) we have
$$
I=-4i\frac{5-\sqrt{20}}{\sqrt{5}\sqrt{5-\sqrt{20}}}\cdot F\left(\arcsin\left(\frac{i}{\sqrt{5-\sqrt{20}}}\right),\frac{5-\sqrt{20}}{5+\sqrt{20}}\right).
$$
A: The answer in terms of elliptic integrals turns out to be very simple – but it requires a very sneaky trick.
As pointed out in Jack's answer, the integrals over $[0,1]$ and $[1,\infty]$ are the same, so
$$I=\int_0^\infty\frac1{\sqrt{x^4+x^3+x^2+x+1}}\,dx=2\int_0^1\frac1{\sqrt{x^4+x^3+x^2+x+1}}\,dx$$
The sneaky trick is to substitute $u=x+\frac1x$, which gives
$$I=2\int_2^\infty\frac1{\sqrt{(u^2-4)(u^2+u-1)}}\,du$$
This changes all the denominator quartic's roots from complex to real, which will make the sequel easier. Furthermore, the roots are very simple: $-2,-\varphi,1/\varphi,2$ where $\varphi$ is the golden ratio. We can now apply Byrd and Friedman 258.00 directly, which says that if $d<c<b<a<y$,
$$\int_a^y\frac1{\sqrt{(t-a)(t-b)(t-c)(t-d)}}\,dt=gF(\psi,m)\\
\text{where }g=\frac2{\sqrt{(a-c)(b-d)}},\psi=\sin^{-1}\sqrt{\frac{(b-d)(y-a)}{(a-d)(y-b)}},m=\frac{(b-c)(a-d)}{(a-c)(b-d)}$$
($m=k^2$ is the parameter, as used by Mathematica and mpmath.) We have $d=-2,c=-\varphi,b=1/\varphi,a=2$, but since $y=\infty$ we have to take the limit of the expression for $\psi$ as $y\to\infty$, which therefore becomes
$$\psi=\sin^{-1}\sqrt{\frac{b-d}{a-d}}=\sin^{-1}\sqrt{\frac{1/\varphi+2}4}=\sin^{-1}\frac\varphi2=\frac{3\pi}{10}$$
The other key numbers follow:
$$g=\frac2{\sqrt{(2+\varphi)(1/\varphi+2)}}=\frac2{\sqrt{4\varphi+3}}$$
$$m=\frac{4(1/\varphi+\varphi)}{(2+\varphi)(1/\varphi+2)}=\frac{4(2\varphi-1)}{4\varphi+3}=8\varphi-12$$
Finally we have the answer in terms of elliptic integrals:
$$\boxed{I=\frac4{\sqrt{4\varphi+3}}F\left(\frac{3\pi}{10},8\varphi-12\right)}$$

#!/usr/bin/env python3
from mpmath import *
mp.dps = 150

print(quad(lambda t: 1/sqrt(polyval((1,1,1,1,1),t)), [0, inf]))

g = 4/sqrt(4*phi+3)
m = 8*phi-12
print(g*ellipf(3*pi/10, m))


P.S. A certain Dr. Sonnhard Graubner gave an answer here, but it's incorrect.
