# Question 268d4

Jan 17, 2018

$0.2854$

#### Explanation:

${\sum}_{n = 0}^{\infty} {r}^{n} = 1 + r + {r}^{2} + {r}^{3} + \ldots = \frac{1}{1 - r}$

Let $r = - {x}^{2}$ and substitute into the series to get:

sum_(n=0)^oo(-x^2)^n=1+(-x^2)+(-x^2)^2+(-x^2)^3+...=1/(1-(-x^2)#

$\to {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n} = 1 - {x}^{2} + {x}^{4} - {x}^{6} + \ldots = \frac{1}{1 + {x}^{2}}$

Now integrate this (term by term):

$\int {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n} \mathrm{dx} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n + 1} / \left(2 n + 1\right)$

$= \int 1 - {x}^{2} + {x}^{4} - {x}^{6} + \ldots \mathrm{dx} = x - {x}^{3} / 3 + {x}^{5} / 5 - {x}^{6} / 6 + \ldots$

$= \int \frac{1}{1 + {x}^{2}} \mathrm{dx} = {\tan}^{- 1} \left(x\right)$

So, we have determined the series for ${\tan}^{-} 1 \left(x\right)$ to be:

${\tan}^{-} 1 \left(x\right) = x - {x}^{3} / 3 + {x}^{5} / 5 - {x}^{7} / 7 + \ldots = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n + 1} / \left(2 n + 1\right)$

It follows that:

$x {\tan}^{-} 1 \left(x\right) = x \left(x - {x}^{3} / 3 + {x}^{5} / 5 - {x}^{7} / 7 + \ldots\right)$

$= {x}^{2} - {x}^{4} / 3 + {x}^{6} / 5 - {x}^{8} / 7 + \ldots = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n + 2} / \left(2 n + 1\right)$

Therefore:

${\int}_{0}^{1} x {\tan}^{-} 1 x \mathrm{dx} = {\int}_{0}^{1} {x}^{2} - {x}^{4} / 3 + {x}^{6} / 5 - {x}^{8} / 7 + \ldots \mathrm{dx} = \int {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n + 2} / \left(2 n + 1\right) \mathrm{dx}$

Integrating term by term gives:

$= {\left[{x}^{3} / 3 - {x}^{5} / 15 + {x}^{7} / 35 - {x}^{9} / 63 + \ldots\right]}_{0}^{1}$

Integrating the expression in summation notation gives.

$= {\left[{\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n + 3} / \left(\left(2 n + 1\right) \left(2 n + 3\right)\right)\right]}_{0}^{1}$

Evaluating the limits, it is clear that this would give:

$= {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} / \left(\left(2 n + 1\right) \left(2 n + 3\right)\right) = \frac{1}{3} - \frac{1}{15} + \frac{1}{35} - \frac{1}{63} + \ldots$

Now that we see the pattern in the series we can just keep adding on until we have reached the 4 decimal place condition:

$\frac{1}{3} = 0.33333$

$\frac{1}{3} - \frac{1}{15} = 0.26667$

$\frac{1}{3} - \frac{1}{15} + \frac{1}{35} = 0.29524$

$\frac{1}{3} - \frac{1}{15} + \frac{1}{35} - \frac{1}{63} = 0.29524$

$\frac{1}{3} - \frac{1}{15} + \frac{1}{35} - \frac{1}{63} = 0.27937$

If we keep going until the value is the same for the 4th decimal place 2 times in a row we get to 52 terms, (note $n = 0$ is the first term so we wish to sum from $n = 0$ to $n = 51$) so:

${\int}_{0}^{1} x {\tan}^{-} 1 x \mathrm{dx} \approx {\sum}_{n = 0}^{51} {\left(- 1\right)}^{n} / \left(\left(2 n + 1\right) \left(2 n + 3\right)\right) = 0.2854$

Analytic evaluation of the integral would yield:

$\frac{1}{4} \left(\pi - 2\right) \approx 0.285398$

so is in agreement with our power series approximation.