Integral Series Interval Of Convergence

You can't evaluate this limit directly, so let's take a look at the behavior of the limit for some easy-to-work-with values of $n$ . Let $n=1.$ The integral is this.

$\begin{aligned} \int\dfrac{x^6}{x^2+1}\text{ }dx&=\int\dfrac{x^6+x^4}{x^2+1}-\dfrac{x^4}{x^2+1}\text{ }dx\\ &=\int x^4-\dfrac{x^4+x^2}{x^2+1}+\dfrac{x^2}{x^2+1}\text{ }dx\\ &=\int x^4-x^2+\dfrac{x^2+1}{x^2+1}-\dfrac{1}{x^2+1}\text{ }dx\\ &=\int x^4-x^2+1-\dfrac{1}{x^2+1}\text{ }dx\\ &=\dfrac{x^5}{5}-\dfrac{x^3}{3}+x-\arctan x^* \end{aligned}$

$\text{ }^*$ Note: there is no need to add a $+C$ because adding a constant does not change the interval of convergence.

Notice the pattern in the coefficients of $x.$ Now expand the integral again with $n=2.$

$\begin{aligned} \int\dfrac{x^{10}}{x^2+1}\text{ }dx&=\int\dfrac{x^{10}+x^8}{x^2+1}-\dfrac{x^8+x^6}{x^2+1}+\dfrac{x^6+x^4}{x^2+1}-\dfrac{x^4+x^2}{x^2+1}+\dfrac{x^2+1}{x^2+1}-\dfrac{1}{x^2+1}\text{ }dx\\ &=\int x^8-x^6+x^4-x^2+1-\dfrac{1}{x^2+1}\text{ }dx\\ &=\dfrac{x^9}{9}-\dfrac{x^7}{7}+\dfrac{x^5}{5}-\dfrac{x^3}{3}+x-\arctan x \end{aligned}$

So there is obviously a pattern in the coefficients of $x.$ The pattern will continue as $x$ tends to $\infty.$ For now, let's ignore the $-\arctan x$ term. Generalize the pattern as a series. $x-\dfrac{x^3}{3}+\dfrac{x^5}{5}-\dfrac{x^7}{7}+\ldots=\lim_{n\rightarrow\infty}\sum_{k=0}^n\dfrac{(-1)^k\times x^{2k+1}}{2k+1}$ Because the summation never has any undefined terms in it, you can eliminate the limit and say this. $x-\dfrac{x^3}{3}+\dfrac{x^5}{5}-\dfrac{x^7}{7}+\ldots=\sum_{k=0}^\infty\dfrac{(-1)^k\times x^{2k+1}}{2k+1}$

This is the power series for $\arctan x.$ Reinserting the $-\arctan x$ term shows that the integral is equal to $0.$

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$\textit{Some things to know before this section}$

$1)\text{ }\displaystyle\lim_{n\rightarrow\infty}a^n$ $\textit{is nonconvergent for}$ $|a|>1$ $\textit{and convergent to}$ $0$ $\textit{for}$ $|a|<1.$

$2)\text{ }\infty-\infty+\infty-\infty+\ldots$ $\textit{is nonconvergent}.$

So since the integral is $0,$ the answer is $(-\infty,\infty)$ right? No, you have to take into consideration the domain of the power series for $\arctan x.$ You need to take into account three ranges: $(-\infty,-1),$ $[-1,1],$ and $(1,\infty).$ Look at the behavior of the power series for $\arctan x$ is for each of these ranges. For $(-\infty,-1)$ and $(1,\infty)$ the magnitudes of the terms increase, so the series will eventually become $\infty-\infty+\infty-\infty+\ldots,$ which is nonconvergent. Now look at $[-1,1].$ The magnitude of the terms gets smaller as the sequence progresses, which is one condition for convergence. You can prove that the power series for $\arctan x$ is convergent for all $x\in[-1,1]$ with the ratio test.

So the first part of the result of the integral is convergent for $x\in[-1,1].$ For the second part, we are looking for the values of $x$ for which $\arctan x$ is defined, which is $(-\infty,\infty).$ The intersection of these sets is the final answer. $(-\infty,\infty)\cap[-1,1]=\boxed{[-1,1]}$ $\text{........................................................................................................................}$ So after all that, the integral is... $0?$ Think about it. Look at the function $\displaystyle\lim_{n\rightarrow\infty}\int\dfrac{x^{4n+2}}{x^2+1}\text{ }dx$ You are finding a formula for finding the area under this curve. When $|x|>1,$ the top part will be an infinite area, resulting in divergence. When $|x|<1,$ it will be $0.$ This results in a convergent area. Now let $x=\pm1.$ The numerator can be arranged like this. $\lim_{n\rightarrow\infty}(\pm1)^{4n+2}=\lim_{n\rightarrow\infty}\left((\pm1)^2\right)^{2n+1}=\lim_{n\rightarrow\infty}1^{2n+1}$ This is an indeterminate exponent of the form $1^\infty,$ but it is equal to $1.$ So the integral is defined and noninfinite at $x=\pm1.$

For $x \in \mathbb{R} - [-1,1]$ , the function clearly diverges, as the numerator increases at a much larger rate, & intuitively, it is not of any use, to divide it with a such smaller number (the denominator), as it appears when $n \to \infty$ . However, for $x \in [-1,1]$ , we have: $x^2 > x^{4n+2}$ , so the function converges to $0$ , and hence it's integral is convergent.

Note that a summation of a function or sequence (basically integrals are also summations) is convergent, iff the function converges to $0$ .

As a follow up, find the interval of convergence of $\displaystyle \lim_{n \to \infty} \int \frac{x^{\frac{4}{n} + 2}}{x^2+1} dx$ .

$\begin{aligned} &\lim\limits_{n\mathop\to\infty}\int\frac{x^{4n+2}}{x^2+1}\,\mathrm{d}x\\ =&\lim\limits_{n\mathop\to\infty}\int\sum_{r\mathop=0}^{2n}(-1)^r x^{2r}-\frac{1}{1+x^2}\,\mathrm{d}x\\ =& x\sum_{r\mathop=0}^{\infty}(-1)^r\frac{ x^{2r}}{2r+1}-\arctan{x}+C \end{aligned}$

If $\displaystyle x\le 1$ then $\left\{ \frac{x^{2r}}{2r+1} \right\}$ decreases monotonically to $0$ , hence by the alternating series test, the series converges. If $\displaystyle x\gt 1$ the limit of the summand does not exist, hence the series diverges.

$\frac{1}{1+x^2}=1-x^2+x^4-x^6+\dots$ . Plug in $x=\pm1$ and you get $1-1+1-1+\dots$ . But, $(\pm1)^{4m+2}=1$ , so $x\neq\pm1$ . The only interval that doesn't include these is $(-1,1)$ .