a question related to integration inequality Let $f:\mathbb R\rightarrow\mathbb R$ be a integrable function such that $\int_{-\infty}^\infty f(t)\,dt = 1$, $\int_{-\infty}^\infty tf(t)\,dt = 0$, $\int_{-\infty}^\infty t^2f(t)\,dt = 1$ and $\int_\lambda^\infty f(t)\,dt \leq e^{-a \lambda^2}$ for some constant $a$. Show that $\int_{1/u}^\infty e^{ut}f(t)\,dt \leq e^{O(u^2)} ,\forall u>0$.
 A: Let the complementary cumulative distribution function of $f(t)$ be
$$\bar{F}(\lambda)=\int_{\lambda}^\infty f(t)dt\leq e^{-a\lambda^2}.$$
Then we can rewrite the integral
$$\int_{1/u}^\infty e^{ut}f(t)dt=-\int_{1/u}^\infty e^{ut}d\bar{F}(t)=-\left.e^{ut}\bar{F}(t)\right|_{1/u}^\infty+u\int_{1/u}^\infty\bar{F}(t)e^{ut}dt.$$
using integration by parts. The first term becomes $e\bar{F}(1/u)$. The $\infty$ end approaches $0$ because $\bar{F}(\lambda)$ is bounded by the Gaussian tail $e^{-a\lambda^2}$. So we have
$$\int_{1/u}^\infty e^{ut}f(t)dt\leq e^{1-a/u^2}+u\int_{1/u}^\infty e^{-at^2+ut\,}dt.$$
In the $u\rightarrow\infty$ limit, the first integral approaches the constant $e$. The second integral approaches (and is bounded by) the incomplete Gaussian integral
$$u\int_0^\infty e^{-at^2+ut\,}dt=u\int_0^\infty e^{-a(t-\frac{u}{2a})^2+\frac{u^2}{4a}}dt\leq u\,e^{\frac{u^2}{4a}}\int_{-\infty}^\infty e^{-a(t-\frac{u}{2a})^2}dt=u\sqrt{\frac{\pi}{a}}e^{\frac{u^2}{4a}},$$
which is then bounded by $e^{Cu^2}$ for any $C>\frac{1}{4a}$ in the $u\rightarrow\infty$ limit.
