# Compute $\lim_{n\rightarrow\infty}\int^n_0\frac1n \sin(nx)e^{-x}dx$

Let $$f_n=\frac{\sin(nx)}{n}e^{-x}$$. My first guess was to try using Monotone Convergence Theorem but we see that $$f_n$$ is not monotone increasing. Note that $$\lim_{n\rightarrow\infty}f_n=0$$ since $$\lim_{n\rightarrow\infty}\frac{\sin(n)}n=0$$.

I would like to try using Dominated Convergence Thm: Let $$g\in L^1(\mathbb{R^n,R})$$, $$g(x)=e^{-x}$$. Then $$|f_n|≤g$$ a.e. We know $$g$$ is integrable, in fact $$\int^\infty_0e^{-x}=1$$. Then by DCT $$\lim_{n\rightarrow\infty}\int f_n=\int0=0$$

Is it correct? I'm not 100% convinced..

• It is almost correct. Take $f_n$ to be $\frac {\sin (nx)} n 1_{(0,n)}(x) e^{-x}$ – Kavi Rama Murthy Oct 21 '20 at 9:33

$$| \int_0^n f_n(x) dx| \le \frac{1}{n}\int_0^n e^{-x}dx=\frac{1}{n}(1-\frac{1}{e^n}) \to 0.$$
Consider $$\int e^{i n x} e^{-x}\,dx=\int e^{i (n+i) x}\,dx=-\frac{i e^{i (n+i) x}}{n+i}$$Now, playing with the complex numbers $$\int \sin(nx)e^{-x}\,dx=-\frac{e^{-x} (\sin (n x)+n \cos (n x))}{n^2+1}$$ $$I_n=\int^n_0\frac1n \sin(nx)e^{-x}\,dx=\frac{1}{n^2+1}-\frac{e^{-n} \left(\sin \left(n^2\right)+n \cos \left(n^2\right)\right)}{n(n^2+1)}$$