difficult limit with a improper integral

Question

It is assigned at my calculus class the following problem.

problem: Evaluate the following limit $$\displaystyle \lim_{n \to \infty} \int \limits_{\frac{1}{(n+1)^2}}^{\frac{1}{n^2}} \frac{e^x\sin^2(x)}{x^{\frac{7}{2}}}\, \mathrm{d}x$$

The problem remains unsolved and the finals are coming soon so I need some help.

Here is my progress:

At first I calculated the $\displaystyle \lim_{x \to 0^{+}} \frac{e^x\sin^2(x)}{x^{\frac{7}{2}}}=\infty$ with the de l' hospital rule.

I thought that if $\displaystyle \lim_{x \to 0^{+}} \frac{e^x\sin^2(x)}{x^{\frac{7}{2}}}<\infty$ then the function $\displaystyle \frac{e^x\sin^2(x)}{x^{\frac{7}{2}}}$ it will be bounded 'near' $0^+$ and as a result $$\displaystyle \lim_{n \to \infty} \int \limits_{\frac{1}{(n+1)^2}}^{\frac{1}{n^2}} \frac{e^x\sin^2(x)}{x^{\frac{7}{2}}}\, \mathrm{d}x=0$$.

Unfortunately $\displaystyle \lim_{x \to 0^{+}} \frac{e^x\sin^2(x)}{x^{\frac{7}{2}}}=\infty$ so how can this problem be solved?

Answer 1

For $\epsilon >0$, find integer $N$ so that if $n>N$ then $1-\epsilon <\frac{\sin ^{2}x}{x^{2}}\leq 1$ if $0<x< \frac{1}{n}$. Then

$\int_{\frac{1}{(n+1)^{2}}}^{\frac{1}{n^{2}}}\frac{e^{x}}{x^{3/2}}\frac{\sin ^{2}x}{x^{2}}dx\geq (1-\epsilon )\int_{\frac{1}{(n+1)^{2}}}^{\frac{1}{n^{2}}}x^{-3/2}dx=2(1-\epsilon )((n+1)-n)=2-\epsilon$.

$\int_{\frac{1}{(n+1)^{2}}}^{\frac{1}{n^{2}}}\frac{e^{x}}{x^{3/2}}\frac{\sin ^{2}x}{x^{2}}dx\leq \int_{\frac{1}{(n+1)^{2}}}^{\frac{1}{n^{2}}}x^{-3/2}dx\leq 2((n+1)-n)=2$.

$\epsilon $ was arbirtray so the limit is $2$.

Answer 2

The integral is less than

$$(1) \,\,\,\,e^{1/n^2}\sin^2(1/n^2)\int_{(n+1)^{-2}}^{n^{-2}} x^{-7/2}\,dx.$$

That last integral equals

$$(2/5)[(n+1)^5 - n^5] = 2n^4 + O(n^3).$$

We know $e^{1/n^2}\sim 1$ and $\sin^2(1/n^2)\sim 1/n^4.$ So the limit in (1) is $2.$ Almost the same argument from below shows a lower bounding function of $n$ with limit $2.$ Therefore the limit in question is $2.$

Answer 3

Consider the integrand and develop it as a Taylor series around $x=0$. You obtain $$\frac{e^x\sin^2(x)}{x^{\frac{7}{2}}}=\frac{1}{x^{3/2}}+\frac{1}{\sqrt{x}}+\frac{\sqrt{x}}{6}-\frac{x^{3/2}}{6}+O\left(x^{5/2}\right)$$ Integrating $$\int\frac{e^x\sin^2(x)}{x^{\frac{7}{2}}}dx=-\frac{2}{\sqrt{x}}+2 \sqrt{x}+\frac{x^{3/2}}{9}-\frac{x^{5/2}}{15}+O\left(x^{7/2}\right)$$ Because of the bounds, we are almost only concerned by the first terms; so, using the first two terms, after some simplifications, $$\int \limits_{\frac{1}{(n+1)^2}}^{\frac{1}{n^2}} \frac{e^x\sin^2(x)}{x^{\frac{7}{2}}}\, \mathrm{d}x\approx \frac{2n(2+n)}{n+1}-\frac{2(n^2-1)}n \approx 2+\frac{2}{n^2}-\frac{2}{n^3}+O\left(\left(\frac{1}{n}\right)^4\right)$$

This had been used for finding the limit but it can be used for evaluating the integral. For example, using $n=10$, numerical integration gives $2.01821$ while the approximation gives $\frac{1009}{500}=2.01800$; using the next term would give $\frac{60547}{30000}\approx 2.01823$