calculus - Determining convergence of improper integrals including $int_{0} ^ {1} frac{lnleft(1+e^xright)-x}{x^2}text{d}x$

Will you please help me figure out whether the following improper integrals converge or not?

1. $$\int _ {0} ^ {\infty} \frac{x^2}{2^x}\text{d}x$$




2. $$\int_{0} ^ {1} \frac{\ln\left(1+e^x\right)-x}{x^2}\text{d}x$$

As for the first one, I have no idea.
As for the second,
I have tried rewriting it as:

$$\int_{0} ^ {1} \frac{\ln\left(\frac{1+e^x}{e^x}\right)}{x^2}\text{d}x$$
but I have no idea if it helps me or not.

Thanks in advance.

Answer

Hint You're on the right track with your manipulation in (2). Since $e^x$ is increasing, for all $x \in [0, 1]$ we have

$$\frac{1 + e^x}{e^x} = 1 + \frac{1}{e^x} \geq 1 + \frac{1}{e} =: C > 0 .$$ Since $\log$ is increasing, we have
$$\frac{\log\left(\frac{1 + e^x}{e^x}\right)}{x^2} \geq \frac{\log C}{x^2}$$ on $(0, 1]$. What can you say about the integral $$\int_0^1 \frac{\log C}{x^2} \, dx$$ relevant to the (direct) comparison test?

For (1), one can determine convergence again by comparing the integrand to a judiciously chosen function.

Additional hint For (1), show that $x^2 \leq a^x$ for all sufficiently large $x$ for some suitable constant $a$.

Remark One can also determine the convergence of the integral in (1) by evaluating it directly, though this is probably slower: Applying integration by parts twice shows that it has value $\frac{2}{(\log 2)^3}$.