Proving $\int_a^\infty \frac{e^{-t}}{t}dt\geq -\log a - 1$

Question

I am trying to prove the following inequaility: $$\int_a^\infty \dfrac{e^{-t}}{t}dt\geq -\log a - 1$$ for all $a$ positive and real. Now, I already try breaking the domain at 1 and using integration by parts but constants appear and the inequalities tend to be of the wrong side. Any help or hint appreciated.

Answer 1

$\int_a^{\infty} \frac {e^{-t}} t \, dt \geq \int_a^{1} \frac {e^{-t}} t \, dt \geq \int_a^{1} \frac {1-t} t \, dt=-log\, a-1+a >-log\, a -1$ for $0<a<1$. For $a\geq 1$ use the fact that $\int_a^{\infty} \frac {e^{-t}} t \, dt +log \, a$ is monotonically inccreasing. (Its derivative is positive)

Answer 2

Integrate by parts to obtain $$ - e^{-a} \log a + \int_a^\infty e^{-t} \log t \, {\rm d}t = -\log a - \gamma + \left\{\left(1 - e^{-a}\right) \log a + \gamma + \int_a^\infty e^{-t} \log t \, {\rm d}t \right\} \, . $$ The expression in the curly bracket vanishes for $a\rightarrow 0$ and the derivative is $$\frac{1-e^{-a}}{a} \geq 0$$ which yields an even better estimate.