Determine divergence of integral using comparison test of $e^{x^2}/x^2$

Question

I'm usually good at determining divergence and using the comparison test, but I can't figure out what function I can use to determine if $$ \int_{0}^{1} \frac{e^{x^2}}{x^2} \, dx$$ is divergent. If anyone can help me, that would be greatly appreciated.

Answer 1

For $x>0$, $e^{x^2}>1$ then $$\dfrac{e^{x^2}}{x^2}>\dfrac{1}{x^2}$$

Answer 2

$$x^2\ge 0\implies e^{x^2}\ge 1\implies \int^1_0\frac{e^{x^2}}{x^2}dx\ge\int^1_0\frac1{x^2}dx=\infty$$

Thus the integral is divergent.

Answer 3

Using the Taylor series: $$\begin{align}\int_{0}^{1} \frac{e^{x^2}}{x^2} \, dx&=\int_{0}^{1} \frac{1+x^2+\frac{x^4}{2!}+O(x^6)}{x^2} \, dx=\\ &=\int_{0}^{1} \frac{1}{x^2}+1+\frac{x^2}{2!}+O(x^4) \, dx>\\ &>\int_{0}^{1} \frac{1}{x^2} \, dx=\\ &=\lim_{x\to 1^-}\left(-\frac1x\right)-\lim_{x\to 0^+}\left(-\frac1x\right)=\\ &=-1+\infty=+\infty. \end{align}$$ Thus, the integral diverges.