# Evaluate and find the limit of an improper integral

I have to make this exercise, where i have to show the following:

$$\frac{f(x)}{x}\rightarrow\infty$$ and $$F(x)\rightarrow\infty$$ for $$x\rightarrow\infty$$

where $$f(x)=e^{x^2}$$ and

$$F(x)=\int_0^x f(t)dt$$. $$(F:\mathbb{R}\rightarrow\mathbb{R})$$

I have showed that the limit is $$\infty$$ for $$\frac{f(x)}{x}$$, when $$x\rightarrow\infty$$

Now i have to show it for $$F(x)$$. I haven't worked with this kind of function before

Do I have to rewrite $$f(x)=e^{x^2}$$ to a taylor-polynomial, and then find the limit of the integral? Or what do I have to do?

It is possible to show from definition that the function diverges. The process looks something like this:

Lemma Suppose that $$f: \mathbb R \to \mathbb R$$ is a function defined and is integrable on interval $$[c,\infty)$$. Suppose also that $$f(x) \gt 0$$ for all $$x$$ in $$[c,\infty)$$. Then if $$\lim_{x \to \infty} f(x) = \infty$$, $$\lim_{x \to \infty}\int_c^x f(t)dt$$ diverges.

Proof Suppose that $$\lim_{x \to \infty}\int_c^x f(t)dt$$ converges to some number $$k$$. Then for any $$\epsilon \gt 0$$, there is a finite number $$N$$ such that $$\lvert k-\int_c^x f(t)dt\rvert \lt \epsilon$$ as long as $$x \gt N$$.

Since $$\lim_{x \to \infty} f(x) = \infty$$, for any finite number $$K$$, there is a finite number $$M$$ such that $$f(x) \gt K$$ as long as $$x \gt M$$.

Now for any $$\epsilon^* \gt 0$$, let $$K^* = 100 \epsilon$$. Therefore there is a finite number $$M^*$$ such that $$f(x) \gt K^*$$ as long as $$x \gt M^*$$, and a finite number $$N^*$$ such that $$\lvert k-\int_c^x f(t)dt\rvert \lt \epsilon^*$$ as long as $$x \gt N^*$$. Let $$N^{**}=\max (M^*,N^*)$$. We have $$f(x) \gt K^*$$ and $$\lvert k-\int_c^x f(t)dt\rvert \lt \epsilon^*$$ as long as $$x \gt N^{**}$$.

However, since $$K^* = 100 \epsilon$$, $$\lvert k-\int_c^{N^{**}+1} f(t)dt\rvert \gt 99 \epsilon$$, which is a contradiction. $$\square$$

Now since $$\lim_{x \to \infty} e^{x^2} = \infty$$, $$\lim_{x \to \infty}\int_0^x e^{t^2}dt$$ diverges. Since $$e^{x^2} \gt 0$$ for all $$x$$ in $$[0,\infty)$$, divergence implies that the function diverges to infinity. Therefore $$F \to \infty$$ as $$x \to \infty$$. $$\square$$

I am quite sure that this answer is sub-optimal and there should be other faster and more convenient ways to do that. :(

just cross check you're indefinite integral, if you still think it's right then https://en.wikipedia.org/wiki/Gaussian_integral. I'm giving you a hint: Think about inequalities for $$f(x)$$

• But my function have positive $x^2$ Commented Mar 29, 2020 at 17:46