Consider the sequence of functions and compute the limit of the integral

Question

Consider the sequence of functions $f_n:[0,1] \to R$ given by

$$f_n(x) = \frac{e^{-(n + x^2)}}{n^2 + x}$$ for $x \in [0,1]$ and $n \geq 1$

Compute $$\lim_{n \to \infty} \int_{0}^{1} f_n(x) dx$$

Maybe I'm overthinking this, but it seems like all I have to do is compute the integral of $f_n(x)$ and then take the limit. However, I am wondering if there is more to it since I can't seem to find an easy way to compute the integral. Can anyone lend me a hand?

Thanks!

Answer 1

I am going to give a solution avoiding any reference to Lebesgue.

First, observe that $f_n(x)\ge 0$ for all $n$ and $x\in[0,1]$, so $$\int_0^1f_n(x)dx\ge 0.$$

Secondly, $f_n(x)\le \frac{1}{n^2+x}$ (observe that $n+x^2>0$ so $e^{-(n+x^2)}\le e^0=1$), so $$\int_0^1f_n(x)dx\le \int_0^1 \frac{1}{n^2+x}dx=\log|n^2+x|\bigg|_0^1=\log(n^2+1)-\log(n^2)=\log(1+1/n^2).$$

Putting al together, $$0\le\int_0^1f_n(x)dx\le\log(1+1/n^2).$$

But $\lim_{n}\log(1+1/n^2)=0$. So the squeeze theorem assures that $$\exists\lim_n \int_0^1 f_n(x)dx=0.$$

Answer 2

$$f_n(x)=\frac{e^{-(n+x^2)}}{n^2+x}=\frac{e^{-n}}{n^2}\frac{e^{-x^2}}{1+x/n^2}$$ now if we make the substitution $u=x/n^2\Rightarrow dx=n^2du$ and our integral becomes: $$I_n=\int\limits_0^{1/n^2}e^{-n}\frac{e^{-n^4u^2}}{1+u}du=e^{-n}\int\limits_0^{1/n^2}\frac{(e^{-u^2})^{n^4}}{1+u}du$$ Now remember that the average of a function is defined as: $$\hat{f}_{[a,b]}=\frac1{b-a}\int_a^bf(x)dx$$ now if $b\to a^+$ it is clear that $\hat{f}\to f(a)$ In our case the range of the interval is tending to zero, in other words: $$\lim_{n\to\infty}n^2\int\limits_0^{1/n^2}\frac{e^{-n^4u^2}}{1+u}du=\lim_{n\to\infty}\lim_{u\to0^+}\frac{e^{-n^4u^2}}{1+u^2}$$ maybe you can use this?