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.

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    $\begingroup$ It doesn't matter for the desired conclusion, but shouldn't it be $\int_0^1e^{-\lambda tx}\:{\rm d}x=\frac{1-e^{-\lambda t}}{\lambda t}$? $\endgroup$ – 0xbadf00d Mar 20 at 13:24
  • $\begingroup$ @Oxbad00d Yes, that was a silly mistake. Thanks for pointing it out. $\endgroup$ – Kavi Rama Murthy Mar 20 at 23:08

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