# prove that $\lim_{x\rightarrow \infty}\int_{x}^{x+a}e^{t^2}dt$ is infinity

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 $$F(x)=\int_{x}^{x+a}f(t)dt \geqslant f(x)$$

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

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 $$a$$.

• 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. Nov 29, 2019 at 20:50

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 $$x \rightarrow \infty$$.