Calculate $\lim_{t\to \infty}\frac{1}{t^2}\int_0^t \ln(e^x+x^2)\,dx$

Question

I need to calculate this: $$ \lim_{t\to \infty}\frac{1}{t^2}\int_0^t \ln(e^x+x^2)\,dx$$

Here's what I've done (in short): Let $f(x)=\ln(e^x+x^2)$, f is continue in $[0, \infty)$, hence for each $t>0$, there is a number $c_t$ such that $\frac{1}{t}\int_0^tf(x)\,dx=f(c_t)$. When $t\to\infty, c_t\to\infty$, so:

$\displaystyle \lim_{t\to \infty}\frac{1}{t}\frac{1}{t}\int_0^t \ln(e^x+x^2)\,dx= \lim_{t\to \infty}\frac{1}{t}f(c_t)=\lim_{t\to \infty}\frac{f(c_t)}{t}=(L'Hospital)\lim_{t\to \infty}\frac{(\ln(e^x+x^2))'}{(t)'}=.....=\lim_{t\to \infty}\frac{e^c_t}{e^c_t}=..=1$

Everyone in my class saying the answer is $1/2$, but I can't find my mistake.

Thanks in advance!

Answer 1

$ G(t)= \int_0^t\ln(e^x+x^2)dx$

$\lim_{t\to \infty}\frac{G(t)}{t^2} = \lim_{t\to \infty}\frac{f(t)}{2t} (L'Hospital)$

the derivative of $t^2$ is $2t$

Answer 2

Let's use L'Hospital rule:

$$ \lim_{t\to \infty}\frac{1}{t^2}\int_0^t \ln(e^x+x^2)\,dx = \lim_{t\to \infty}\frac{1}{2t} \ln(e^t+t^2) ={1\over 2}\lim_{t\to \infty}\frac{e^t+2t}{e^t+t^2} ={1\over 2}$$

Answer 3

Loosely speaking, one expects the $x^2$ perturbation to be negligible for $t\to+\infty$, and from $$ \frac{1}{t^2}\int_{0}^{t}\log(e^x+x^2)\,dx \stackrel{?}{\approx}\frac{1}{t^2}\int_{0}^{t}\log(e^x)\,dx = \frac{1}{t^2}\int_{0}^{t}x\,dx = \frac{1}{2} $$ $\frac{1}{2}$ is definitely the most reasonable answer. One just needs to turn the $\stackrel{?}{\approx}$ above into a rigorous thing,
i.e. to prove that $$ \lim_{t\to +\infty}\frac{1}{t^2}\int_{0}^{t}\log(1+x^2 e^{-x})\,dx = 0. $$ On the other hand $x^2 e^{-x}$ is a non-negative and bounded function on $\mathbb{R}^+$, hence $\int_{0}^{t}\log(1+x^2 e^{-x})\,dx=O(t)$ and the previous identity is trivial. We do not even need derivatives.

Answer 4

By the monotonicity of the $\log$: $$0\le x\le t\implies x\le\log(e^x + x^2)\le\log(e^x + t^2) = x + \log(1 + t^2e^{-x})\le x + \log(1 + t^2).$$ Integrating: $$\frac{t^2}2\le\int_0^t\log(e^x + x^2)\,dx\le\frac{t^2}{2} + t\log(1 + t^2).$$ Dividing by $t^2$ and squeezing, we conclude $$\lim_{t\to\infty}\int_0^t\log(e^x + x^2)\,dx = \frac12.$$