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$$\lim_{n\to\infty}\int_\frac{1}{n}^n\frac{\arctan x} {(x^2+x+1)}$$

Could you help me find this limit? Any ideas are helpful.

jvdhooft
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Octavian Alixandrescu
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1 Answers1

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Set $x=\dfrac1y$

$$I=\int_{1/n}^n\dfrac{\arctan x}{x^2+x+1}dx=\int_n^{1/n}\dfrac{\arctan\dfrac1y}{\dfrac1{y^2}+\dfrac1y+1}\left(-\dfrac{dy}{y^2}\right)=\int_{1/n}^n\dfrac{\arctan\dfrac1y}{1+y+y^2}dy$$

$$I+I=\int_{1/n}^n\dfrac{\arctan x+\arctan\dfrac1x}{x^2+x+1}dx$$

Using Are $\mathrm{arccot}(x)$ and $\arctan(1/x)$ the same function?,

for $x>0,$ $$2I=\int_{1/n}^n\dfrac{\dfrac\pi2}{x^2+x+1}dx$$

lab bhattacharjee
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  • Thank you, really helped me. At a previous exercise I proved that arctan(x) +arctan(1/x)=pi/2 so that actually helped me. And also I had to find the indefinite integral of $$\frac{1}{x^2+x+1}$$ so in the end it was pretty easy to solve IT. Thank you once again! – Octavian Alixandrescu Apr 28 '20 at 09:20