Can anyone help me calculate this integral ? I thought about using the dominant convergence theorem but the region is always changing so im not sure i can just do it and i dont see any other way

$$\lim_{n\to\infty}\int_{0}^{n} \frac{1}{1+x^{2n}}\,\mathrm{d}x$$



A standard approach to problems of this type would be to write the integral as follows $$ \int_0^n \frac{1}{1+x^{2n}}\,\mathrm{d} x =\int_0^{\infty} \frac{1}{1+x^{2n}}\cdot \mathbb{1}_{[0,n]}(x)\,\mathrm{d} x $$ and then apply the DCT on the functions $$ f_n(x) = \frac{1}{1+x^{2n}}\cdot \mathbb{1}_{[0,n]}(x).$$ Edit: I chose to answer only the difficult part of the question. Due to some comments which make clear that the conclusion is not clear from here, I'll specify further.

So, the difficult part is: how to apply the DCT on a changing interval? I converted this problem to a standard integral where the DCT needs to be applied to a fixed interval.

The pointwise limit can be calculated as $$f(x) = \lim_{n\to\infty} f_n(x) = \begin{cases} 1 & \text{if}\ x\in[0,1) \\ 0.5 & \text{if}\ x=1 \\ 0 & \text{if}\ x>1. \end{cases} $$ From here, it follows that if DCT is applicable then we could calculate $$ \lim_{n\to\infty} \int_0^n \frac{1}{1+x^{2n}}\,\mathrm{d} x = \lim_{n\to\infty} \int_0^{\infty} f_n(x)\,\mathrm{d} x = \int_0^{\infty} \lim_{n\to\infty}f_n(x)\,\mathrm{d} x \\ = \int_0^{\infty} f(x)\,\mathrm{d} x = \int_0^{1} 1,\mathrm{d} x = 1. $$ To apply DCT, we need to find an integrable function which dominates $f_n(x)$ for all $n\in\mathbb{N}$. Note that the function which is equal to $1$ dominates all $f_n$, but this is not integrable on $[0,\infty)$. A standard way to go is to split up the proof in two cases, $x<1$ and $x\geq1$.

I will leave it up to OP two fill in all details, but you should end up with something as follows. For $0\leq x<1$, $$ |f_n(x)| = \left|\frac{1}{1+x^{2n}}\cdot \mathbb{1}_{[0,n]}(x)\right| =\frac{1}{1+x^{2n}} \leq \frac{1}{1+1}=\frac{1}{2} $$ For $x>1$, $$ |f_n(x)| = \left|\frac{1}{1+x^{2n}}\cdot \mathbb{1}_{[0,n]}(x)\right| \leq \frac{1}{1+x^{2n}} \leq \frac{1}{1+x^2} $$ Note that $1/2$ is integrable over $[0,1)$ and $1/(1+x^2)$ is integrable over $[1,\infty)$ (it has the $\arctan$ as antiderivative and it integrates to $\pi/4$). This means the DCT is applicable and that is why the above calculation of the limit holds.

  • $\begingroup$ Alright that makes sense, but what is the function that dominates all $f_n$ in $[0,\infty]$ and i also got calculate that integral wich to me isnt obvious, Can i consider the function constant 1 and the integral is always finite in [0,n] and it is n ? $\endgroup$ – Pedro Santos Apr 15 at 11:36
  • $\begingroup$ I don't think is a proper way to answer the question. A dominating function has to be specified. @Stan Tendijck $\endgroup$ – Kavi Rama Murthy Apr 15 at 12:01
  • $\begingroup$ @PedroSantos Let $f(x)=1$ for $x<1$ and $f(x)=\frac 1 {1+x^{2}}$ for $x \geq 1$. Then $f$ is integrable and $|f_n| \leq f$ for all $n$. $\endgroup$ – Kavi Rama Murthy Apr 15 at 12:04
  • $\begingroup$ Yeah thats true but how do i do the limit how do i know what is the integral of $f_n(x)$? $\endgroup$ – Pedro Santos Apr 15 at 12:15
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    $\begingroup$ I just converted my answer to a full solution instead of tackling only the changing interval part. I hope this is more clear to you. $\endgroup$ – Stan Tendijck Apr 15 at 12:23

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