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Can I use the Taylor series to find this integrals? $$ \lim_{n\to \infty } \int _{0}^n \frac{\arctan x}{x^2 + x + 1}dx $$ $$\lim _{n\to \infty }n \int_{-1}^0(x + e^x)^{n}dx = \: ?$$

Lola
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  • you mean to approximate the integral? It depends: if the radius of convergence of some Taylor series cover the range of integration yes, if not no. To evaluate numerically some integral the standard and easiest method is the generalized Simpson's rule. –  May 07 '17 at 16:33
  • One thing I was thinking of to do literal long division of the series of $arctanx$ by $x^2+2x+1$. Got a couple of terms $x-2x^2+(8/3)x^3...$, problem is that the interval of convergence of the $arctan$ is limited, so there is a problem. Why not a numerical method? – imranfat May 07 '17 at 16:34
  • I never heard of that rule, I want a simple approach to it since I'm still in highschool and just started working with integrals – Lola May 07 '17 at 16:35
  • What do you mean by numerical method? – Lola May 07 '17 at 16:35
  • I edited the first limit its x not 2x ;the answer in my book is $$\boxed{\dfrac{\pi^2}{6\sqrt{3}}} $$ – Lola May 07 '17 at 16:41
  • @Lola This coincides with my PARI/GP-calculations, the other limit seems to be $\frac{1}{2}$ – Peter May 07 '17 at 16:44
  • Yes thats the answer for the other limit – Lola May 07 '17 at 16:50
  • i think your result is wrong – Dr. Sonnhard Graubner May 07 '17 at 17:32

1 Answers1

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Here is a relatively simple standard approach to evaluating the first of your integrals.

Let $$I = \int_0^\infty \frac{\arctan x}{x^2 + x + 1} \, dx.$$ Dividing up the interval of integration as follows $$I = \int_0^1 \frac{\arctan x}{x^2 + x + 1} \, dx + \int_1^\infty \frac{\arctan x}{x^2 + x + 1} \, dx.$$ In the second of the integrals appearing to the right, enforcing a substitution of $x \mapsto 1/x$, one has \begin{align} I &= \int_0^1 \frac{\arctan x}{x^2 + x + 1} \, dx + \int_0^1 \frac{\arctan \left (\frac{1}{x} \right )}{x^2 + x + 1} \, dx\\ &= \int_0^1 \frac{\arctan (x) + \arctan \left (\frac{1}{x} \right )}{x^2 + x + 1} \, dx\\ &= \frac{\pi}{2} \int_0^1 \frac{dx}{x^2 + x + 1} \tag1\\ &= \frac{\pi}{2} \int_0^1 \frac{dx}{\left (x + \frac{1}{2} \right )^2 + \frac{3}{4}}\\ &= \frac{\pi}{\sqrt{3}} \left [\arctan \left (\frac{2x + 1}{\sqrt{3}} \right ) \right ]_0^1\\ &= \frac{\pi^2}{6 \sqrt{3}} \end{align}

At step (1) the following well-known result for the inverse tangent function has been used: $$\arctan (x) + \arctan \left (\frac{1}{x} \right ) = \frac{\pi}{2}, \qquad x > 0.$$

omegadot
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