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I have to calculate

$$\int_0^\infty \frac{n\sin x}{1+n^2x^2}\,\mathrm{d}x$$

and I would like to use the DCT. I have said that

$$\forall x\in (0,1)\qquad\frac{n\sin x}{1+n^2 x^2} \leq \frac{n\sin x}{2 n x} \leq \frac{n x}{2 n x} \leq \frac{1}{2}$$

and

$$\forall x\in (1,\infty)\qquad\frac{n\sin x}{1+n^2x^2}\leq\frac{1}{2}.$$

My questions are the following:

  1. Is my estimation in $(0,1)$ right?

  2. I think that is right because I am in $(0,1)$ and I am dividing $x$ over $x$ and it is well defined. Is this right?

  3. Why don't I have to count the point $x=0$? I mean the integral is defined over $0$ and $\infty$ so it has $0$ in its domain so my estimation $\frac{n\sin x}{1+n^2 x^2}\leq\frac{1}{2}$ is not right because I am diving $\frac{0}{0}$?

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  • $\begingroup$ Is $\lim_{n \to \infty}\int_0^\infty \frac{n\sin x}{1+n^2x^2}\,dx$ what you want to calculate? $\endgroup$
    – Martin R
    Apr 15 at 13:04
  • $\begingroup$ See math.stackexchange.com/q/3789989/42969 $\endgroup$
    – Martin R
    Apr 15 at 13:10
  • $\begingroup$ @MartinR. If you have time to waste, have a look at my answer in the post you gave the link of. Cheers :-) $\endgroup$ Apr 15 at 14:43

2 Answers 2

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Yes, your estimate is correct, as

$$0\leq(nx-1)^2=n^2x^2-2nx+1$$

implies that

$$1+n^2x^2\geq2nx,$$

and the rest of the estimates are clear. Now the reason why you can ignore the point $x=0$ is because the set $\{0\}$ has Lebesgue measure $0$, so in particular,

$$\int_{[0,\infty)}f(x)\,\mathrm{d}x=\underbrace{\int_{\{0\}}f(x)\,\mathrm{d}x}_{=0}+\int_{(0,\infty)}f(x)\,\mathrm{d}x=\int_{(0,\infty)}f(x)\,\mathrm{d}x,$$

meaning that the integral is the same if you discount the origin.

Now since your goal was to compute the integral using the DCT, I will point out another problem with your solution. While your bounds are correct, to be able to apply the DCT you want to bound your sequence by an integrable function, however $\frac{1}{2}$ is not integrable on $(0,\infty)$, as

$$\int_0^\infty\frac{1}{2}\,\mathrm{d}x=\infty.$$

Thus the DCT is not applicable with this bound. What you could do instead is the following:

$$\left\lvert\frac{n\sin x}{1+n^2x^2}\right\rvert\leq\frac{n}{1+n^2x^2}\leq\frac{n^2}{1+n^2x^2}=\frac{1}{\frac{1}{n^2}+x^2}\leq\frac{1}{x^2}$$

for all $x\in[1,\infty)$ and $n\geq1$, and the function

$$x\mapsto\begin{cases} \frac{1}{2},&x\in(0,1), \\ \frac{1}{x^2},&x\in[1,\infty) \end{cases}$$

is clearly integrable on $(0,\infty)$, and so by the DCT, as

$$\lim_{n\to\infty}\int_0^\infty\frac{n\sin x}{1+n^2x^2}\,\mathrm{d}x=\int_0^\infty\lim_{n\to\infty}\frac{n\sin x}{1+n^2x^2}\,\mathrm{d}x$$

(I'll leave computing this to you).

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  • $\begingroup$ best answer ever $\endgroup$
    – nathan
    Apr 15 at 14:15
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Just for your curiosity.

If, for your problem, you use the same method as here, you will obtain $$I_n=\int\limits_{0}^{\infty}\frac{n \sin (x)}{1+n^2x^2}\,dx=\text{Shi}\left(\frac{1}{n}\right) \cosh \left(\frac{1}{n}\right)-\text{Chi}\left(\frac{1}{n}\right) \sinh \left(\frac{1}{n}\right)$$

Asymptotically $$I_n=\frac{\log (n)-\gamma +1}{n}+\frac{6 \log (n)-6 \gamma +11}{36 n^3}+O\left(\frac{1}{n^5}\right)$$ which is very accurate even for small values $n$.

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