Question

0%7Bx%7D%5E%7B2%7D%20%20%2B%201%7D%7B%20%7Bx%7D%5E%7B4%7D%20%20%2B%20%20%7Bx%7D%5E%7B2%7D%20%20%2B%201%7D%20%20%5Cright%29%20%5Cleft%28%20%20%5Cfrac%7B%20ln%20%5Cleft%281%20-%20x%20%2B%20%20%7Bx%7D%5E%7B2%7D%20%20-%20%20%7Bx%7D%5E3%20%2B%20%20%5Cdots%20%2B%20%20%7Bx%7D%5E%7B2020%7D%20%20%20%5Cright%29%20%7D%7B%20ln%28x%29%20%7D%20%5Cright%29%20%5C%3A%20dx" id="TexFormula1" title=" \displaystyle \rm\int_{0}^{ \infty } \left( \frac{ {x}^{2} + 1}{ {x}^{4} + {x}^{2} + 1} \right) \left( \frac{ ln \left(1 - x + {x}^{2} - {x}^3 + \dots + {x}^{2020} \right) }{ ln(x) } \right) \: dx" alt=" \displaystyle \rm\int_{0}^{ \infty } \left( \frac{ {x}^{2} + 1}{ {x}^{4} + {x}^{2} + 1} \right) \left( \frac{ ln \left(1 - x + {x}^{2} - {x}^3 + \dots + {x}^{2020} \right) }{ ln(x) } \right) \: dx" align="absmiddle" class="latex-formula">​

Answer 1

Recall the geometric sum,

$\displaystyle \sum_{k=0}^{n-1} x^k = \frac{1-x^k}{1-x}$

It follows that

$1 - x + x^2 - x^3 + \cdots + x^{2020} = \dfrac{1 + x^{2021}}{1 + x}$

So, we can rewrite the integral as

$\displaystyle \int_0^\infty \frac{x^2 + 1}{x^4 + x^2 + 1} \frac{\ln(1 + x^{2021}) - \ln(1 + x)}{\ln(x)} \, dx$

Split up the integral at x = 1, and consider the latter integral,

$\displaystyle \int_1^\infty \frac{x^2 + 1}{x^4 + x^2 + 1} \frac{\ln(1 + x^{2021}) - \ln(1 + x)}{\ln(x)} \, dx$

Substitute $x\to\frac1x$ to get

$\displaystyle \int_0^1 \frac{\frac1{x^2} + 1}{\frac1{x^4} + \frac1{x^2} + 1} \frac{\ln\left(1 + \frac1{x^{2021}}\right) - \ln\left(1 + \frac1x\right)}{\ln\left(\frac1x\right)} \, \frac{dx}{x^2}$

Rewrite the logarithms to expand the integral as

$\displaystyle - \int_0^1 \frac{1+x^2}{1+x^2+x^4} \frac{\ln(x^{2021}+1) - \ln(x^{2021}) - \ln(x+1) + \ln(x)}{\ln(x)} \, dx$

Grouping together terms in the numerator, we can write

$\displaystyle -\int_0^1 \frac{1+x^2}{1+x^2+x^4} \frac{\ln(x^{2020}+1)-\ln(x+1)}{\ln(x)} \, dx + 2020 \int_0^1 \frac{1+x^2}{1+x^2+x^4} \, dx$

and the first term here will vanish with the other integral from the earlier split. So the original integral reduces to

$\displaystyle \int_0^\infty \frac{1+x^2}{1+x^2+x^4} \frac{\ln(1-x+\cdots+x^{2020})}{\ln(x)} \, dx = 2020 \int_0^1 \frac{1+x^2}{1+x^2+x^4} \, dx$

Substituting $x\to\frac1x$ again shows this integral is the same over (0, 1) as it is over (1, ∞), and since the integrand is even, we ultimately have

$\displaystyle \int_0^\infty \frac{1+x^2}{1+x^2+x^4} \frac{\ln(1-x+\cdots+x^{2020})}{\ln(x)} \, dx = 2020 \int_0^1 \frac{1+x^2}{1+x^2+x^4} \, dx \\\\ = 1010 \int_0^\infty \frac{1+x^2}{1+x^2+x^4} \, dx \\\\ = 505 \int_{-\infty}^\infty \frac{1+x^2}{1+x^2+x^4} \, dx$

We can neatly handle the remaining integral with complex residues. Consider the contour integral

$\displaystyle \int_\gamma \frac{1+z^2}{1+z^2+z^4} \, dz$

where γ is a semicircle with radius R centered at the origin, such that Im(z) ≥ 0, and the diameter corresponds to the interval [-R, R]. It's easy to show the integral over the semicircular arc vanishes as R → ∞. By the residue theorem,

$\displaystyle \int_{-\infty}^\infty \frac{1+x^2}{1+x^2+x^4}\, dx = 2\pi i \sum_\zeta \mathrm{Res}\left(\frac{1+z^2}{1+z^2+z^4}, z=\zeta\right)$

where $\zeta$ denotes the roots of $1+z^2+z^4$ that lie in the interior of γ; these are $\zeta=\pm\frac12+\frac{i\sqrt3}2$. Compute the residues there, and we find

$\displaystyle \int_{-\infty}^\infty \frac{1+x^2}{1+x^2+x^4} \, dx = \frac{2\pi}{\sqrt3}$

and so the original integral's value is

$505 \times \dfrac{2\pi}{\sqrt3} = \boxed{\dfrac{1010\pi}{\sqrt3}}$