Show that there is \(\varepsilon>0\) such that if \(0<|x|<\varepsilon\) then \[ \int_0^x e^{t^2}\,dt>\frac{x^2}{2}+\ln(1+x). \]
Solution
1. Expand both sides near the origin.
For \(x\) near \(0\), \[ e^{t^2}=1+t^2+O(t^4), \] so integration gives \[ \int_0^x e^{t^2}\,dt =x+\frac{x^3}{3}+O(x^5). \] Also, \[ \ln(1+x) =x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+O(x^5), \] and therefore \[ \frac{x^2}{2}+\ln(1+x) =x+\frac{x^3}{3}-\frac{x^4}{4}+O(x^5). \] Subtracting, \[ g(x):=\int_0^x e^{t^2}\,dt-\frac{x^2}{2}-\ln(1+x) =\frac{x^4}{4}+O(x^5). \]
2. Deduce strict positivity on a punctured neighborhood.
There exist \(C>0\) and \(0<\delta<1\) such that \[ \left|g(x)-\frac{x^4}{4}\right|\le C|x|^5 \qquad(|x|<\delta). \] Choose \[ \varepsilon:=\min\left(\delta,\frac{1}{8C}\right). \] Then for \(0<|x|<\varepsilon\), \[ g(x) \ge \frac{x^4}{4}-C|x|^5 \ge \frac{x^4}{8}>0. \] Hence \[ \boxed{ \int_0^x e^{t^2}\,dt>\frac{x^2}{2}+\ln(1+x) \quad\text{for }0<|x|<\varepsilon.} \] At \(x=0\) the two sides are equal, which is why the puncture is necessary.