Trick 1: Rewrite the integral so it has the bounds you want.
\begin{equation}
\int_0^\infty \nu d \nu = \int_0^1\nu d \nu + \int_1^\infty \nu d \nu = \frac{1}{2} + \int_1^\infty \nu d \nu
\end{equation}
Trick 2: Multiply the integrand and summand by $1=\lim_{\epsilon\rightarrow 0} e^{- \epsilon \nu}$
\begin{equation}
\Delta E = \frac{\pi}{L} \left(\sum_{\nu=1}^\infty\left[\lim_{\epsilon\rightarrow 0}e^{-\epsilon \nu}\right]\nu -\int_1^\infty \left[\lim_{\epsilon\rightarrow 0}e^{-\epsilon \nu}\right] \nu d \nu - \frac{1}{2}\right)
\end{equation}
Trick 3: Take the limit outside the sum/integral (this is not an obvious step and I suspect might not really be fully rigorous, but ok at a physics level of rigor)
\begin{equation}
\Delta E = \frac{\pi}{L} \left[\lim_{\epsilon\rightarrow 0} \left(\sum_{\nu=1}^\infty e^{-\epsilon \nu}\nu -\int_1^\infty e^{-\epsilon \nu} \nu d \nu \right) \right]- \frac{\pi}{2L}
\end{equation}
Trick 4: We apply the Euler-Maclaurin formula to evaluate the difference in the parentheses.
Let's write out the first few terms of the general formula
\begin{equation}
\sum_{i=a}^b f(i) - \int_a^b f(x) dx = \frac{f(a)+f(b)}{2} + \frac{f'(b) - f'(a)}{12} - \frac{f'''(a) - f'''(b)}{720} + \cdots
\end{equation}
where in our case, $f(x)=e^{-\epsilon x} x$, $a=1$, and $b=\infty$.
The first term yields
\begin{equation}
\frac{f(a)+f(b)}{2} = \frac{1}{2}
\end{equation}
This exactly cancels the $-\pi/2L$ term in $\Delta E$ above.
For the second term, we need the derivative
\begin{equation}
f'(x) = e^{-\epsilon x} - \epsilon e^{-\epsilon x} x
\end{equation}
We will ignore the second term, since in the end we want to take the limit $\epsilon\rightarrow 0$, and that term will not contribute. Therefore,
\begin{equation}
\frac{f'(b) - f'(a)}{12} = -\frac{1}{12} e^{-\epsilon} + {\rm term\ that\ vanishes\ as\ \epsilon\rightarrow 0}
\end{equation}
In the limit $\epsilon\rightarrow 0$, this simply becomes $-1/12$.
Finally, let's consider the third term. We have
\begin{equation}
f'''(x) = -\epsilon^2 e^{-\epsilon x}\left(\epsilon x - 3\right)
\end{equation}
Since this will go to zero as $\epsilon \rightarrow 0$, we can neglect this term. Similarly, all higher derivatives of $f$ (which appear in higher order terms in the Euler-Maclaurin formula) will go to zero as $\epsilon\rightarrow 0$.
Putting this together:
\begin{eqnarray}
\Delta E &=& \frac{\pi}{L} \left[\lim_{\epsilon\rightarrow 0} \left(\sum_{\nu=1}^\infty e^{-\epsilon \nu}\nu -\int_1^\infty e^{-\epsilon \nu} \nu d \nu \right) \right]- \frac{\pi}{2L} \\
&=& \left(\frac{\pi}{2L} - \frac{\pi}{12 L} + 0 \right) - \frac{\pi}{2L} \\
&=& -\frac{\pi}{12 L}
\end{eqnarray}
as desired.
