Assuming that $\lambda_1,\ldots,\lambda_n\in\mathbb{R}^+$ are the eigenvalues, we are just claiming:
$$\sum_{k=1}^{n}\left(1-\frac{1}{\lambda_k}\right) \leq \sum_{k=1}^{n}\log(\lambda_k)\leq \sum_{k=1}^{n}\left(\lambda_k-1\right)\tag{1}$$
or, by setting $\lambda_k=e^{u_k}$:
$$\sum_{k=1}^{n}\left(1-e^{-u_k}\right) \leq \sum_{k=1}^{n} u_k \leq \sum_{k=1}^{n}\left(e^{u_k}-1\right)\tag{2}$$
that trivially follows from the convexity of the exponential function.
If $$I_m=\int_0^{\infty} \exp \left[ - \frac{(x - \mu)^2}{2 \sigma^2} \right] x^m \, dx$$
$$\frac{I_m}{2^{\frac{m-1}{2}} \sigma ^m }=\sqrt{2} \mu \Gamma \left(\frac{m+2}{2}\right) \,
_1F_1\left(\frac{1-m}{2};\frac{3}{2};-\frac{\mu ^2}{2 \sigma
^2}\right)+$$ $$\sigma \Gamma \left(\frac{m+1}{2}\right) \,
_1F_1\left(-\frac{m}{2};\frac{1}{2};-\frac{\mu ^2}{2 \sigma
^2}\right)$$ where appears Kummer confluent hypergeometric function.
Let $\mu=\sqrt{2} \sigma t$
$$\frac{I_m}{2^{\frac{m-1}{2}} \sigma ^{m+1} }=2 \,t\, \Gamma \left(\frac{m}{2}+1\right) \,
_1F_1\left(\frac{1-m}{2};\frac{3}{2};-t^2\right)+$$ $$\Gamma \left(\frac{m+1}{2}\right) \,
_1F_1\left(-\frac{m}{2};\frac{1}{2};-t^2\right)$$
Not proved, it seems that the second term is slightly larger than the first term.
$$\frac{I_m}{2^{\frac{m-1}{2}} \sigma ^{m+1} }< 2\,\Gamma \left(\frac{m+1}{2}\right) \,
_1F_1\left(-\frac{m}{2};\frac{1}{2};-t^2\right)$$
For example, using $\mu=3$, $\sigma=1$ and $m=5$, the exact value is $1398.699006$ while the bound is $1398.699435$.
At this point, I am stuck.
Best Answer
Applying integration by parts $n$ times gives
\begin{align*} \int_{a}^{b} \frac{e^x}{x} \, \mathrm{d}x = e^b R_n(b) - e^a R_n(a) + \int_{a}^{b} \frac{n! e^{x}}{x^{n+1}} \, \mathrm{d}x, \end{align*}
where $R_n(x) = \sum_{k=1}^{n} \frac{(k-1)!}{x^k}$. Now, as $a, b, (b/a) \to \infty $, the last integral is bounded by $\mathcal{O}(e^b / b^{n+1})$, and so, we get
$$ \int_{a}^{b} \frac{e^x}{x} \, \mathrm{d}x = e^b R_{n}(b) + \mathcal{O}(e^b /b^{n+1}). $$
as $a, b, (b/a) \to \infty$, for each fixed $n \geq 1$.