# Asymptotics of a function

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

• can you show that the most important contributions come from a very small neighbourhood of $0_+$? If yes, this is a standard application of Laplace method, yielding $I_n\sim n^{-1}$ – tired Mar 28 '17 at 17:34
• I've only been told that finding the asymptotics of a sequence $(a_n)_{n\geq 1}$ means providing another sequence $(b_n)_{n\geq 1}$ such that $\lim_{n\to\infty} a_n/b_n =1$ – user430196 Mar 28 '17 at 17:37

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 $u(x)=\text{sinc}(x),\,v'(x)=(1+x)^{-n}$

• This is really helpful. However, I would like to concentrate on showing it in through the integral and ending up with $\lim_{n\to\infty} I_n n= 1$. Apparently there is a substitution Im missing – user430196 Mar 28 '17 at 18:05
• what do you mean by "showing it in through the integral "? – tired Mar 28 '17 at 18:06
• @Taozi this is the same for big $n$ ;-) – tired Mar 28 '17 at 18:16
• @tired You are very right, just that the difference in Mathematica's figure makes me want to say something :-) – Taozi Mar 28 '17 at 18:18
• (+1) back at you. One note; the error is more acurately $O(n^{-3})$ – Mark Viola Mar 28 '17 at 18:31

Let $I_n$ be given by

\begin{align} I_n&=\int_0^1 \frac{\sin(x)}{x(1+x)^n}\,dx\\\\ &=\int_0^1 (1+x)^{-n}\,dx+\sum_{k=1}^\infty\frac{(-1)^{k}}{(2k+1)!}\int_0^1x^{2k}(1+x)^{-n}\,dx\\\\ &=\frac{1}{n-1}-\frac{2}{2^n(n-1)}+\sum_{k=1}^\infty\frac{(-1)^{k}}{(2k+1)!}\int_0^1 x^{2k}(1+x)^{-n}\,dx\tag 1 \end{align}

We can evaluate the integral on the right-hand side of $(1)$ be making the substitution $x\to x-1$ and using the binomial theorem to write $(x-1)^{2k}=\sum_{\ell = 0}^{2k}\binom{2k}{\ell}(-1)^{2k-\ell}x^\ell$. Proceeding we find

$$\int_0^1x^{2k}(1+x)^{-n}\,dx=\sum_{\ell = 0}^{2k}\binom{2k}{\ell}(-1)^{2k-\ell}\frac{1-2^{-n+\ell+1}}{n-(\ell+1)}$$

which shows that $\int_0^1x^{2k}(1+x)^{-n}\,dx=O\left(\frac{1}{n^{2k+1}}\right)$

Finally, we have

$$I_n=\frac{1}{n-1}-\frac{1}{3(n-1)(n-2)(n-3)}+O\left(\frac1{n^5}\right)$$

• the canonical way (+1) – tired Mar 28 '17 at 18:29
• @tired Yes. I was going to approach this differently, but thought "What the heck?" – Mark Viola Mar 28 '17 at 18:31
• i was going to employ Laplace method but decided later on that something more elementary seems to be in order – tired Mar 28 '17 at 18:32