TITLE: Asymptotics of a function
QUESTION [6 upvotes]: I have to find the asymptotics of the following integral
$$\int^1_0 \frac{\sin(x)}{x(1+x)^n} dx$$
as $n\to\infty$.
I know I am supposed to use the lebesgue dominated convergence theorem and create a sequence $b_n$ and aim for $\lim_{n\to\infty} a_n/b_n =1$ but I can't seem to get my head round it.
Any help I would be very much grateful for

REPLY [7 votes]: Denote the integral in question by $I_n$ and squeeze it baby!
Notice that the inequalities $x-x^3/3!<\sin(x)<x$ hold on $x\in(0,1)$
so
$$
\int_0^1dx\frac{1-x^2/3!}{(1+x)^n}<I_n<
\int_0^1dx\frac{1}{(1+x)^n}
$$
doing the algebra (which is elementary) you get
$$
\frac{1}{n-1}+\mathcal{O}(n^{-2})<I_n<\frac{1}{n-1}+\mathcal{O}( n^{-2})
$$
which shows that $I_n\sim n^{-1}$ or $\lim_{n\rightarrow\infty}nI_n=1$

For the record: 
Integration by parts also works nicely here: use $u(x)=\text{sinc}(x),\,v'(x)=(1+x)^{-n}$