I am trying to find whether $\int_0^\infty \frac{\ln(x^2)}{e^{x^2}} \,\mathrm{d}x$ is convergent or divergent.
Firstly, I broke the integral up into two improper integrals, namely:
- the integral from 0 to 1;
- the integral from 1 to infinity
and then finding their corresponding limits.
I am trying to use the Comparison Test, and was notified by someone that the following worked:
on $(0,1)$ the exact value of $\ln(x) < \frac{1}{\sqrt{x}}$ AND exact value of $e^{-x^2} \le 1$.
on $(1,\infty)$, exact value of $\ln(x)<x$ and $x<2e^{x/2}$ AND $e^{-x^2}<e^{-x}$.
I am confused on why they are taking the exact value and how they could imply the above inequalities.
However, I do know how to prove for $(1,\infty)$ that e^{-x^2}
\begin{align*} x=1 &\implies x^2>x \\ &\implies e^{x^2}>e^{x} \\ &\implies \frac{1}{e^x} > \frac{1}{e^{x^2}} \end{align*}
but the rest I am so unsure on.
Can someone please help me prove this and explain the reason for the exect values?
According to Mathematica the integral is convergent and gives: $$\int_0^{\infty} \frac{\log \left(x^2\right)}{e^{x^2}} \mathbb dx = -\frac{1}{2} \sqrt{\pi } (\gamma +\log (4)).$$