# How can we show that $\left|\frac{e^{-\lambda t}-1}t\right|\le|\lambda|$ for all $t\in\mathbb R\setminus\left\{0\right\}$ and $\lambda\in\mathbb R$?

How can we show that $$\left|\frac{e^{-\lambda t}-1}t\right|\le|\lambda|$$ for all $$t\in\mathbb R\setminus\left\{0\right\}$$ and $$\lambda\in\mathbb R$$?

Clearly, $$e^{-\lambda t}\in[0,1]$$ for all $$t,\lambda\ge0$$, but that doesn't help.

The inequality is false. See what happens as $$\lambda \to -\infty$$ with $$t=1$$ to see that it is false. [Indeed, when $$\lambda=-n$$ and $$t=1$$ the inequality says $$|e^{n}-1| \leq n$$ which is false as you can see from the series expansion of $$e^{n}$$]. It is true for $$\lambda, t \geq 0$$. For $$\lambda =0$$ this is obvious. For $$\lambda > 0$$ note that $$\int_0^{1} e^{-\lambda tx} dx=\frac {1-e^{-\lambda t}} {\lambda t}$$. Use the fact that the integrand is $$\leq 1$$. The inequality is also true when both parameters are negative.
• It doesn't matter for the desired conclusion, but shouldn't it be $\int_0^1e^{-\lambda tx}\:{\rm d}x=\frac{1-e^{-\lambda t}}{\lambda t}$? – 0xbadf00d Mar 20 at 13:24