Evaluate and find the limit of an improper integral

Question

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?

Thanks in advance.

Answer 1

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 $f(x)$

Answer 2

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. :(