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Exercise: Compute $$\lim_{n\to\infty}\int_\limits{0}^{1}\frac{nx}{1+n^2x^4}dx.$$

Solution:$$\int_\limits{0}^{1}\frac{nx}{1+n^2x^4}dx=\frac{1}{n}\int_\limits{0}^{n}\frac{y}{1+\frac{y^4}{n^2}}dy=n\int_\limits{0}^{n}\frac{y}{n^2+y^4}dy=\frac{n}{2}\int_\limits{0}^{n}\frac{dy^2}{n^2+y^4}=\frac{n}{2}\int_\limits{0}^{n^2}\frac{du}{n^2+u^2}$$ $$=\frac{1}{2}\int_\limits{0}^{n^2}\frac{\frac{du}{n}}{1+(\frac{u}{n})^2}=\frac{1}{2}\arctan(n)$$

$\lim_\limits{n\to\infty}\dfrac{1}{2}\arctan(n)=\dfrac{\pi}{4}$

Questions:

$1$) I examined this solution and I did not understand the following step: $$n\int_\limits{0}^{n}\frac{y}{n^2+y^4}dy=\frac{n}{2}\int_\limits{0}^{n}\frac{dy^2}{n^2+y^4}$$ How is this possible? Is there any theorem behind it? If there is can you direct me to a proof of it.

$2$) If $f_n$ converges uniformly on $[a,b]$ then $$\int_\limits{a}^{b}f(x) dx=\lim_\limits{n\to\infty}\int_\limits{a}^{b}f_n(x)$$

I computed $\lim_\limits{n\to\infty}\frac{nx}{1+n^2x^4}=0$ so the function converges uniformly, is that right? Why cannot I apply the the given result?

Thanks in advance!

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  • $\begingroup$ 1) To me this seems just like substitution. Say $n>0$ and $x=y^2$ then $y>0$ and for every $x$ there is one $y$. $dx/dy=2y$, so you have $$n \int_0^n\frac{y}{n^2+y^4}dy=n\int_0^{n^2}\frac{x}{n^2+x^2}\frac{1}{2\sqrt{x}}dx=\frac{n}{2}\int_0^{n^2}\frac{\sqrt{x}}{n^2+x^2}dx=\frac{n}{2}\int_0^{n^2} \frac{y}{n^2+y^4}dy^2.$$ The last step is trivial. $\endgroup$ – ty. Mar 26 '18 at 14:52
  • $\begingroup$ By definition $\exists\lim_{n\to\infty}\frac{nx}{1+n^2x^4}$ means pointwise convergence. $\endgroup$ – Martín-Blas Pérez Pinilla Mar 26 '18 at 17:12
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For 1, $2ydy=d(y^2)$. It is used to change the variable from $y$ to $y^2$.

For 2, the limit is not 0, but a Dirac delta function. You can go to one of the intermediate steps, where you have $$\frac{n/2}{n^2+u^2}$$ If you look for example on wikipedia, You can write the Dirac delta function as a limit of a sequence of functions. In particular, the Poisson kernel $$\frac{1}{\pi}\frac{\epsilon}{\epsilon^2+x^2}$$ looks exactly like your function to integrate.

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  • $\begingroup$ Please show me point 2. Why does the limit define a Dirac delta function for $x\in[0,1]$? $\endgroup$ – Pedro Gomes Mar 26 '18 at 15:08
  • $\begingroup$ How much is $0\cdot\infty$? $\endgroup$ – Andrei Mar 26 '18 at 15:50
  • $\begingroup$ I'll add the links to my reply $\endgroup$ – Andrei Mar 26 '18 at 15:58
  • $\begingroup$ Thanks I got it. However I am still stuck at point 1). How do you arrive at $2ydy=d(y^2)$? I am using second derivative but I get nothing like that. $\endgroup$ – Pedro Gomes Mar 26 '18 at 16:03
  • $\begingroup$ What's the first derivative of $y^2$? Just do a substitution $u=y^2$. Then $du=2ydy$. $\endgroup$ – Andrei Mar 26 '18 at 16:09
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For 2, you simply computed the pointwise limit of the integrand. Why do you believe that the mode of convergence improves to uniform convergence? After all, the proof shows that the convergence cannot be uniform. This can also be read out from the graph of $y = \frac{nx}{1+n^2x^4}$ for $n = 1, \cdots, 100$:

$\hspace{3.5em}$ graphs

This tells that the 'total mass' concentrates toward the origin, hence the vanishing of pointwise density does not correctly represents the limit of the total mass.

This kind of behavior can be resolved once you look at the integrand in the correct scale. In this case, you can apply the substitution $y = \sqrt{n}x$ to check that

$$ \int_{0}^{1} \frac{nx}{1+n^2x^4} \, dx = \int_{0}^{\sqrt{n}} \frac{y}{1+y^4} \, dy \xrightarrow[n\to\infty]{} \int_{0}^{\infty} \frac{y}{1+y^4} \, dy = \frac{\pi}{4}. $$

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