Indication: $\displaystyle \lim_{x \to \infty}{\int^{x}_{1}{\ln(f(t))dt}}$

Question

Let $f$ be a function $f:\mathbb{R}\to\mathbb{R}$, $f(x)=4x^3 + 1$.

I have to find out $\displaystyle \lim_{x \to \infty}{\int^{x}_{1}{\ln(f(t))dt}}$.

My intuition is telling me the limit is $+\infty$. What I tried is was to rewrite as: $$\int^{x}_{1}{\ln(f(t))dt}=\int^{x}_{1}{t' \cdot\ln(f(t))dt}=t\cdot \ln(f(t))|^{x}_{0}-\int^{x}_{1}{\frac{t\cdot f'(t)}{f(t)}dt}.$$ The first member is going to $+\infty$ and now I have to focus over the second member which can be written as $ \displaystyle \int^{x}_{1}{\frac{12t^3}{4t^3 +1}dt}=\int^{x}_{1}{3dt}-\int^{x}_{1}{\frac{3}{4t^3+1}dt}$. Again, the firs member goes to $+\infty$.

Answer 1

$$\int^{x}_{1}\ln(4t^3+1)\textrm{d}t \ge \int^{x}_{1}\ln(4t^3)\textrm{d}t =\int^{x}_{1}\ln(4)+\ln (t^3)\textrm{d}t \ge \int^{x}_{1}\ln(4)\textrm dt= (\ln 4)(x-1)$$

Answer 2

You can figure out how fast the integral converges to infinity with the estimates

$$\int_1^x \log 4 t^3 d t \le \int_1^x \log (4 t^3 + 1)\,dt \le \int_1^x \log 5 t^3 dt$$

and
$$\int_1^x \log ( a x^3) \, dt = \log a \cdot (x-1) + 3 (x \log x -x + 1) $$