2
$\begingroup$

Evaluate the following: $\displaystyle\lim_{n\to\infty} \displaystyle\int_0^\infty \dfrac{n\sin x}{1+n^2x^2}\ dx$

Attempt: My idea to use the Dominated Convergence Theorem. It can easily be shown that $f_n(x)=\dfrac{n\sin x}{1+n^2x^2}$ converges pointwise to $0$ on $\left(0,\infty\right)$ but I'm having difficulties on trying to get an integrable function $g$ that would satisfy $|f_n|\leq g$. I tried using the inequalities $|\sin t|\leq t$ for small $t$ and of course the obvious one which is $|\sin t|\leq 1$ but the function I get is not integrable. Any help would be great.

$\endgroup$
3
  • 3
    $\begingroup$ Hint. Substitute $u = nx$. $\endgroup$ – Sangchul Lee May 5 '19 at 0:38
  • 1
    $\begingroup$ Hint, use $u=nx$. $\endgroup$ – Jo' May 5 '19 at 0:39
  • $\begingroup$ Thank you very much! $\endgroup$ – John Thompson May 5 '19 at 0:42
5
$\begingroup$

Hint: Let $u=nx$, then your integral becomes \begin{align} \int^\infty_0 \frac{\sin\frac{u}{n}}{1+u^2}\ du \end{align}

$\endgroup$
1
  • $\begingroup$ Oh thank you very much! $\endgroup$ – John Thompson May 5 '19 at 0:42
4
$\begingroup$

Just added for your curiosity.

We can compute the antiderivative, starting with $$\frac n {1+n^2x^2}=\frac n {(nx-i)(nx+i)}=\frac {ni} 2 \left(\frac 1 {nx+i}-\frac 1 {nx-i} \right)$$ which make that we face the problem of $$I=\int \frac{\sin(x)}{x+a}\,dx$$ Let $x+a=t$ to make $$I=\int \frac{\sin(t-a)}{t}\,dt=\cos(a)\int \frac{\sin(t)}{t}\,dt-\sin(a)\int \frac{\cos(t)}{t}\,dt $$ that is to say $$I=\cos (a)\, \text{Si}(t)-\sin (a) \,\text{Ci}(t)$$ where appear the sine and cosine integrals (don' worry : sooner or later, you will leran about them).

Using the integration bounds and then the asymptotics, you will end with $$I_n=\int_0^\infty \dfrac{n\sin x}{1+n^2x^2}\, dx=\frac{\log \left({n}\right)-\gamma +1}{n}+O\left(\frac{1}{n^3}\right)$$ which, for sure, shows the limit but also how it is approached. Moreover, as shown in the table below, this gives an approximation of the result $$\left( \begin{array}{ccc} 1 & 0.422784 & 0.646761 \\ 2 & 0.557966 & 0.599204 \\ 3 & 0.507132 & 0.521764 \\ 4 & 0.452270 & 0.459176 \\ 5 & 0.406444 & 0.410274 \\ 6 & 0.369091 & 0.371446 \\ 7 & 0.338385 & 0.339943 \\ 8 & 0.312778 & 0.313865 \\ 9 & 0.291112 & 0.291902 \\ 10 & 0.272537 & 0.273130 \\ 20 & 0.170926 & 0.171014 \\ 30 & 0.127466 & 0.127495 \\ 40 & 0.102792 & 0.102804 \\ 50 & 0.086696 & 0.086703 \\ 60 & 0.075286 & 0.075290 \\ 70 & 0.066733 & 0.066735 \\ 80 & 0.060060 & 0.060062 \\ 90 & 0.054696 & 0.054697 \\ 100 & 0.050280 & 0.050281 \end{array} \right)$$

$\endgroup$
-1
$\begingroup$

This is not a complete answer:

The integral from Jacky Chong can be evaluated with Mathematica: $$\int_0^{\infty } \frac{\text{Sin}\left[\frac{u}{n}\right]}{1+u^2} \ \, du=\frac{\sqrt{\pi } \text{MeijerG}\left[\{\{0\},\{\}\},\left\{\{0,0\},\left\{-\frac{1}{2}\right\}\right\},\frac{1}{4n^2}\right]}{4 n}$$ Therefore the solution can be expressed in form of a MeijerG-Function and can be further analyzed by an asymptotic series expansion. This leads to the same result as from Claude Leibovici.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.