Is $\int_{\infty}^{\infty^2} f(t) dt$ always $0$?

Question

Let $f(t)$ be Riemann-integrable and $\int_{-\infty}^{\infty} f(t) dt=1$.

If I now look at the limit $\lim\limits_{x \rightarrow \infty}{F(x)}$, where

$$F(x)=\int_{x}^{x^2} f(t) dt$$

I wonder if this limit always has to be $0$ or if it is possible that $\lim\limits_{x \rightarrow \infty}{F(x)} \neq 0$ ?

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In my opinion this is not possible, since $f(t)$ has to be bounded by definition, thus it has to converge for $x \rightarrow \pm\infty$.

However I'd like to know if the fact, that my Integral goes from $\infty$ to "$\infty^2$" could make some trouble.

Answer 1

Yes, it has to be $0$. Let $G(x)=\int_{-\infty}^xf(t)\,\mathrm dt$. Then $\lim_{x\to\infty}G(x)=1$. But$$\lim_{x\to\infty}\int_x^{x^2}f(t)\,\mathrm dt=\lim_{x\to\infty}G(x^2)-G(x)=1-1=0.$$

Answer 2

If $\int_{-\infty}^{\infty} f(t) \,dt=1$ then by definition there exists $c \in \mathbb{R}$ such that $$\lim_{x\to-\infty}\int_{x}^{c} f(t) \,dt + \lim_{x\to\infty}\int_{c}^{x} f(t) \,dt= \int_{-\infty}^{c} f(t) \,dt + \int_{c}^{\infty} f(t) \,dt =1$$

In particular these two limits exist.

Now we have

$$\lim_{x\to\infty} \int_{x}^{x^2} f(t) \,dt = \lim_{x\to\infty} \int_{c}^{x^2} f(t) \,dt - \lim_{x\to\infty} \int_{c}^{x} f(t) \,dt = \int_{c}^{\infty} f(t) \,dt - \int_{c}^{\infty} f(t) \,dt =0$$

Answer 3

Yes,

$$F(x)=\int_{x}^{x^2} f(t) dt = \int_{0}^{x^2} f(t) dt -\int_{0}^{x} f(t) dt$$

Both the integrals on the RHS converge to $\int_{0}^{\infty} f(t) dt$ as $x\to \infty$