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This is the limit:

$$\lim_{h\rightarrow 0}\dfrac{ \displaystyle\int^{\pi+he^{-\frac{1}{h}}}_{0}x^2e^{-x^2}dx-\int^{\pi}_{0}x^2e^{-x^2}dx}{he^{-\frac{1}{h}}}$$

In the site I've seen it proposed to calculate, two solutions have been shown, one of them (the more straightforward one) dismissed as incorrect and the other one accepted. I see more dubious the accepted proof, and this makes more ashtonishing for me the rejecting of the other.

I write the calculations proposed for you to judge. And my questions are: is the dismissal correct? and, if affirmative, why?

1.-

$$\lim_{h\rightarrow 0}\dfrac{\int^{\pi+he^{-\frac{1}{h}}}_{0}x^2e^{-x^2}dx-\int^{\pi}_{0}x^2e^{-x^2}dx}{he^{-\frac{1}{h}}} = \displaystyle \lim_{h\rightarrow 0}\dfrac{\int^{\pi+he^{-\frac{1}{h}}}_{\pi}x^2e^{-x^2}dx }{he^{-\frac{1}{h}}}$$

Making $x = \pi+h\,e^{-\frac{1}{h}}\,t \;\;\longrightarrow\;\;\text{dx} = h\,e^{-\frac{1}{h}}\,\text{dt}$

$$\lim_{h\to 0}\dfrac{\int^{\pi+he^{-\frac{1}{h}}}_{\pi}x^2e^{-x^2}dx }{he^{-\frac{1}{h}}} =\lim_{h\to 0}\int_{0}^{1} \left( \pi + he^{-\frac{1}{h} \, t}\right)^2 e^{-\left( \pi + he^{-\frac{1}{h} \, t}\right)^2} \text{dt}=$$

$$=\int_{0}^{1} \lim_{h\to 0}\left( \pi + he^{-\frac{1}{h} \, t}\right)^2 e^{-\left( \pi + he^{-\frac{1}{h} \, t}\right)^2}\text{dt} = \pi^2 e^{-\pi^2}$$

2.-

Let be $F(t)=\int_{0}^{t}x^2e^{-x^2}dx$, then $F'(t)=t^2e^{-t^2}$

Let be $\Delta t=+he^{-\frac{1}{h}}$, then $h\rightarrow{0}\Rightarrow{}\Delta t\rightarrow{0}$

$$\lim_{h\rightarrow 0}\dfrac{\int^{\pi+he^{-\frac{1}{h}}}_{0}x^2e^{-x^2}dx-\int^{\pi}_{0}x^2e^{-x^2}dx}{he^{-\frac{1}{h}}}=\lim_{\Delta t\rightarrow 0}\dfrac{\int^{\pi+\Delta t}_{0}x^2e^{-x^2}dx-\int^{\pi}_{0}x^2e^{-x^2}dx}{\Delta t} =\lim_{\Delta t \to 0}{}\frac{F(\pi+\Delta t)-F(\pi)}{\Delta t}=F'(\pi)=\pi^2e^{-\pi^2}$$

Rafa Budría
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  • just to clarify, which one is the "incorrect" one? – John Feb 17 '17 at 17:18
  • :) Well, I've left this point open. But if you are uncomfortable I can tell you wich one. I don't pretend make a riddle. – Rafa Budría Feb 17 '17 at 17:22
  • Note that $h\to 0$ and $\Delta t\to 0$ are NOT equivalent. You must require $h>0$ for this to be the case (i.e., you are only considering $h\searrow 0$). – MPW Feb 17 '17 at 17:24
  • @MPW Yes, I considered this point, but, hasn't the other proof the same problem? Aren't them both restricted to $ h>0$? And why is possible that "integral sign jump"? – Rafa Budría Feb 17 '17 at 17:27
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    The first proof passes the limit under the integral sign. This isn't always allowed and needs justification. – Trevor Norton Feb 17 '17 at 17:27
  • http://math.stackexchange.com/questions/253696/can-a-limit-of-an-integral-be-moved-inside-the-integral – Trevor Norton Feb 17 '17 at 17:28
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    Obviously the second method is simpler and better and preferred. The first one unnecessarily complicates the problem by relying on interchange of limit operation with the integral sign. – Paramanand Singh Feb 18 '17 at 07:28

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