1

$$\lim_{n\rightarrow\infty}\int_0^1\frac{ne^x}{1+n^2x^2} dx$$ I deduced that the sequence of function is not uniformly convergent to it's functional limit i.e. $f(x)=0$ but couldn't proceed further to calculate the final value. Can someone please guide me on how to proceed?

Ani
  • 19
  • Also: https://math.stackexchange.com/q/1679031/42969, https://math.stackexchange.com/q/3947283/42969, https://math.stackexchange.com/q/3899853/42969 – all found with Approach0 – Martin R Dec 12 '21 at 16:34

1 Answers1

1

Hint :

  1. Show that$$\int_0^1\frac{ne^x}{1+n^2x^2} dx = \int_0^{+\infty}\frac{e^{x/n}}{1+x^2} \chi_{[0,n]} dx $$

  2. Show that for every $x \geq 0$,

$$\left| \frac{e^{x/n}}{1+x^2} \chi_{[0,n]} \right| \leq \frac{e}{1+x^2}$$

  1. Use the dominated convergence theorem to conclude that $$\lim_{n \rightarrow +\infty} \int_0^1\frac{ne^x}{1+n^2x^2} dx = \frac{\pi}{2}$$
TheSilverDoe
  • 29,720
  • @MartinR IMO, this is not a proper duplicate (I searched for them this time). In the linked questions, there is an unknown function $f$ in the integral, which makes the question more complicated than this one. I agree that the links give several answers to the OP's question, but in the special case where $f = \exp$, the answer can be really more elementary, and I guess such an answer has its place on this website. – TheSilverDoe Dec 12 '21 at 16:39
  • Well, I disagree. This for example does not look far more complicated to me: https://math.stackexchange.com/a/1679034/42969. – Martin R Dec 12 '21 at 16:44
  • @MartinR Ok. I admit I did not find this particular answer, so I did not come across any answer using Lebesgue's DCT. That's why I thought my answer could be useful. – TheSilverDoe Dec 12 '21 at 16:46