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Let $a>0$ be a real number, prove that $\lim_{x\rightarrow \infty}\int_{x}^{x+a}e^{t^2}dt$ is infinity.

What I tried:

$\int_{x}^{x+a}e^{t^2}dt \geqslant m(x+a-x) = ma = e^{x^2}a $

Where $m = inf\{e^{t^2}|t \in [x,x+a]\}$

Then I wanted to take the limit of $e^{x^2}a $ but I'm not sure if I can do that since $a$ might be any number, especially a very small number.

Also, I tried writing $\int_{x}^{x+a}e^{t^2}dt = \int_{0}^{x+a}e^{t^2}dt - \int_{0}^{x}e^{t^2}dt$ but got stuck.

The official solution which I didn't understand is:

Using the monotonic property of the integral and the monotonic property of $e^{x^2}$ we get:

$\int_{x}^{x+a}e^{t^2}dt \geqslant e^{x^2} \overset{x\rightarrow \infty}{\rightarrow} \infty$

My questions are:

  1. Can I go on with my solution and write that $\lim_{x\rightarrow \infty}e^{x^2}a = \infty?(a>0)$, if yes, why can I do that? I mean couldn't $a \rightarrow 0$ and then we the limit of zero times infinity? (sorry for the bad phrasing, I didn't learn math using English).

  2. What did they do in the official solution, how did they know that $\int_{x}^{x+a}e^{t^2}dt \geqslant e^{x^2}$? and can I say that for every monotonic function $f(x),a>0$ happens that $F(x)=\int_{x}^{x+a}f(t)dt \geqslant f(x)$

Thanks in advance!

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  • $\begingroup$ The limits of the integral should be not the same as the variables of the integral $\endgroup$ Nov 29, 2019 at 20:18
  • $\begingroup$ You should write $$\int_{y}^{y+a}e^{x^2}dx$$ $\endgroup$ Nov 29, 2019 at 20:19
  • $\begingroup$ $$\lim_{y\to \infty}\int_{y}^{y+a}e^{x^2}dx$$ $\endgroup$ Nov 29, 2019 at 20:22
  • $\begingroup$ Sorry, I didn't notice the title was wrong since it doesn't show up under the text editor. $\endgroup$
    – Titan3
    Nov 29, 2019 at 20:26

2 Answers 2

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Since $a$ is a positive constant, it can't tend to zero any more than it makes sense for $1/2\to 0$. Due to this, your solution is correct. A positive constant times a function that tends to $\infty$ will tend to $\infty$.

The official solution, as written, is wrong. It would also need the factor of $a$.

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  • $\begingroup$ I got so frustrated trying to derive and get to what's in the official solution that I started to forget the basics and question myself. Thank you very much. $\endgroup$
    – Titan3
    Nov 29, 2019 at 20:50
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MVT for integrals:

Let $a>0;$

$\displaystyle{\int_{x}^{x+a}}e^{t^2}dt =e^{s^2}\int_{x}^{x+a}1dt=$

$e^{s^2}a \ge ae^{x^2}.$

Recall $s \in [x,x+a]$, and $e^{t^2}$ is an increasing function of $t$

Take the limit $x \rightarrow \infty$.

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