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

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:

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

