It's an exercise I took from an exam of Measure Theory.

Using all the theorems it needs, evaluate: $$\lim_{n \rightarrow \infty} \int_0^{+\infty} \frac{e^{-n^2x}}{\sqrt{|x-n^2|}} dx $$

It has an additional hint: from $n>2$ look at the integral like $$ \int_0^{+\infty} = \int_0^1 + \int_1^{n^2-n} + \int_{n^2-n}^{n^2+n}+\int_{n^2+n}^{+\infty}$$

Moreover the function has no elementary primitive, but it is the product of two functions that do have elementary primitive. This can be helpful after having done the right considerations.

  • $\begingroup$ I suppose you mean "calculate" or "find". You cannot "solve" a limit, limit is not a problem or question. $\endgroup$ – Alexey Aug 31 '19 at 8:38
  • $\begingroup$ yes, i'm not english mother-tongue so excuse me for the mistake $\endgroup$ – Gabrielek Aug 31 '19 at 8:40
  • $\begingroup$ Not mine either :). $\endgroup$ – Alexey Aug 31 '19 at 11:37

Following the given hint we have that for $n\geq 2$,

i) if $x\in [0,1]$ then $$\frac{e^{-n^2x}}{\sqrt{|x-n^2|}}\leq \frac{1}{\sqrt{n^2-1}},$$ ii) if $x\in [1,n^2-n]$ then $$\frac{e^{-n^2x}}{\sqrt{|x-n^2|}}\leq \frac{e^{-n^2}}{\sqrt{n}},$$ if $x\in [n^2+n,+\infty)$ then $$\frac{e^{-n^2x}}{\sqrt{|x-n^2|}}\leq \frac{e^{-n^2x}}{\sqrt{n}}.$$ By using these estimates, you may show that the three related integrals go to zero. It remains to consider the integral: $$\int_{n^2-n}^{n^2+n}\frac{e^{-n^2x}}{\sqrt{|x-n^2|}} \,dx\leq e^{-n^2(n^2-n)}\int_{n^2-n}^{n^2+n}\frac{dx}{\sqrt{|x-n^2|}}=e^{-n^2(n^2-n)}2\int_{0}^{n}\frac{dt}{\sqrt{t}}.$$ Can you take it from here?

  • $\begingroup$ for the last integral can't i say that it's less than $\frac{e^{-n^4+n^2}}{\sqrt{n}}(2n)$ ? $\endgroup$ – Gabrielek Aug 31 '19 at 8:47
  • $\begingroup$ Yes! Note that in $[n^2-n,n^2+n]$ the denominator $\sqrt{|x-n^2|}$ goes to zero. $\endgroup$ – Robert Z Aug 31 '19 at 8:51

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