# Evaluating the limit of the given integral

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.

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

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

• Oh thank you very much! – John Thompson May 5 '19 at 0:42

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 $$\left( \begin{array}{ccc} 1 & 0.422784 & 0.646761 \\ 2 & 0.557966 & 0.599204 \\ 3 & 0.507132 & 0.521764 \\ 4 & 0.452270 & 0.459176 \\ 5 & 0.406444 & 0.410274 \\ 6 & 0.369091 & 0.371446 \\ 7 & 0.338385 & 0.339943 \\ 8 & 0.312778 & 0.313865 \\ 9 & 0.291112 & 0.291902 \\ 10 & 0.272537 & 0.273130 \\ 20 & 0.170926 & 0.171014 \\ 30 & 0.127466 & 0.127495 \\ 40 & 0.102792 & 0.102804 \\ 50 & 0.086696 & 0.086703 \\ 60 & 0.075286 & 0.075290 \\ 70 & 0.066733 & 0.066735 \\ 80 & 0.060060 & 0.060062 \\ 90 & 0.054696 & 0.054697 \\ 100 & 0.050280 & 0.050281 \end{array} \right)$$
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.