Join the IDNLearn.com community and get your questions answered by knowledgeable individuals. Get accurate and detailed answers to your questions from our knowledgeable and dedicated community members.
Sagot :
Recall the geometric sum,
[tex]\displaystyle \sum_{k=0}^{n-1} x^k = \frac{1-x^k}{1-x}[/tex]
It follows that
[tex]1 - x + x^2 - x^3 + \cdots + x^{2020} = \dfrac{1 + x^{2021}}{1 + x}[/tex]
So, we can rewrite the integral as
[tex]\displaystyle \int_0^\infty \frac{x^2 + 1}{x^4 + x^2 + 1} \frac{\ln(1 + x^{2021}) - \ln(1 + x)}{\ln(x)} \, dx[/tex]
Split up the integral at x = 1, and consider the latter integral,
[tex]\displaystyle \int_1^\infty \frac{x^2 + 1}{x^4 + x^2 + 1} \frac{\ln(1 + x^{2021}) - \ln(1 + x)}{\ln(x)} \, dx[/tex]
Substitute [tex]x\to\frac1x[/tex] to get
[tex]\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}[/tex]
Rewrite the logarithms to expand the integral as
[tex]\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[/tex]
Grouping together terms in the numerator, we can write
[tex]\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[/tex]
and the first term here will vanish with the other integral from the earlier split. So the original integral reduces to
[tex]\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[/tex]
Substituting [tex]x\to\frac1x[/tex] again shows this integral is the same over (0, 1) as it is over (1, ∞), and since the integrand is even, we ultimately have
[tex]\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[/tex]
We can neatly handle the remaining integral with complex residues. Consider the contour integral
[tex]\displaystyle \int_\gamma \frac{1+z^2}{1+z^2+z^4} \, dz[/tex]
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,
[tex]\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)[/tex]
where [tex]\zeta[/tex] denotes the roots of [tex]1+z^2+z^4[/tex] that lie in the interior of γ; these are [tex]\zeta=\pm\frac12+\frac{i\sqrt3}2[/tex]. Compute the residues there, and we find
[tex]\displaystyle \int_{-\infty}^\infty \frac{1+x^2}{1+x^2+x^4} \, dx = \frac{2\pi}{\sqrt3}[/tex]
and so the original integral's value is
[tex]505 \times \dfrac{2\pi}{\sqrt3} = \boxed{\dfrac{1010\pi}{\sqrt3}}[/tex]
We appreciate your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. For dependable answers, trust IDNLearn.com. Thank you for visiting, and we look forward to helping you again soon.
