Analytically, how is $\int_0^{\infty} t^{-1}e^{-t} \geq \frac{1}{e} \int_0^1 \frac{dt}{t}$?

Question

I found this inequelity in a proof that uses a lower bound for $\Gamma(0)$ being the integral $\frac{1}{e} \int_0^1 \frac{dt}{t}$ which tends to $+\infty$ and I want to verify it analytically because I don't see how this bound came, knowing that $\int_0^{\infty} t^{-1}e^{-t}$ cannot be expressed using usual primitives, I tried using $\int_0^{\infty} \int_x^{\infty} t^{-1}e^{-t}dtdx$ but it didn't lead me to this lower bound integral.

Answer 1

Well, your function is positive, so

$$ \int_0^{\infty} t^{-1} e^{-t}\textrm{d}t\geq \int_0^1 t^{-1}e^{-t}\textrm{d}t\geq \int_0^1t^{-1} e^{-1}\textrm{d}t, $$ where we've also used that $e^{-1}\leq e^{-t}$ for every $t\in [0,1]$ and that $t^{-1}$ is positive.