The following lemma is from Titchmarsh's The Theory of the Riemann Zeta-Function:
I have difficulties in getting both the estimates:
1- $O((\sum_1^{\infty} n^{-{\sigma}} e^{-\delta n})^2) = O((\int_1^{\infty} x^{-{\sigma}} e^{-\delta x})^2)$, so how to estimate the integral or any other way to get $O({\delta}^{2 \sigma -2})$?
2- $O(\sum_1^{\infty} n^{1-{2\sigma}} e^{-\delta n} \sum_1^{n/2} 1/r )= O(\sum_1^{\infty} n^{1-{2\sigma}} e^{-\delta n} \log n) = O(\int_1^{\infty} x^{1-{2\sigma}} e^{-\delta x} \log x )$, so how to estimate the integral or any other way to get $O({\delta}^{2 \sigma -2} \log \dfrac{1}{\delta})$?
WolframAlpha couldn't be useful.
Both of these estimates rely on the same basic idea, which is that after a basic substitution you get the bound multiplied by some integral that obviously converges to some value independent of $\delta$.
I'll look at the first one. The second is essentially the same.
As you've noticed, the first estimate reduces to studying the (square of the) integral
$$ \int_1^\infty t^{-\sigma} e^{-\delta t} dt = \int_1^\infty t^{1 - \sigma} e^{-\delta t} \frac{dt}{t}. $$ Perform the change of variables $t \mapsto \tfrac{t}{\delta}$ to see that this is $$ \int_\delta^\infty (t/\delta)^{1 - \sigma} e^{-t} \frac{dt}{t} = \delta^{\sigma - 1} \int_\delta^\infty t^{1 - \sigma} e^{-t} \frac{dt}{t}. $$ The behavior of the integral on the right depends on what $\sigma$ and $\delta$ are. If $\sigma < 1$, then the integral converges $$ \int_\delta^\infty t^{1 - \sigma} e^{-t} \frac{dt}{t} \leq \int_0^\infty t^{1 - \sigma}{e^{-t}} \frac{dt}{t} = \Gamma(1 - \sigma) < \infty. $$ If $\delta \geq 1$, then also have $$ \int_\delta^\infty t^{1 - \sigma} e^{-t} \frac{dt}{t} \leq \int_1^\infty t^{1 - \sigma}e^{-t} \frac{dt}{t} = O_\sigma(1).$$ But if $\delta$ is allowed to be arbitrarily small and $\sigma \geq 1$, then more care is necessary.
I presume that the bound is for fixed $\sigma$ as $\delta \to \infty$, in which case the two simple cases above are sufficient. This isn't clear from the image.
If desired, it is possible to give more precise bounds. The integral $$ \int_\delta^\infty t^{1 - \sigma} e^{-t} \frac{dt}{t} = \Gamma(1 - \sigma, \delta) $$ is the "upper incomplete gamma function". Very good bounds and approximations are known. I wouldn't expect these to be necessary for whatever application Titchmarsh has in mind.
The conclusion is that if either $\delta \geq 1$ or if $\sigma < 1$, then $$ \int_1^\infty t^{-\sigma} e^{-\delta t} dt = \delta^{\sigma - 1} O_\sigma(1). $$
A similar change-of-variables and basic bounding applies for the second sum (and with the same regions of where these simple bounds work).