Euler-Maclaurin formula for Casimir Effect I’ve been reading this http://www.hep.caltech.edu/~phys199/lectures/lect5_6_cas.pdf for a derivation of the Casimir Effect. On page 4 we see that
$$\Delta{E} = \frac{\pi}{L}\left(\sum_{\nu = 1}^{\infty}\nu - \int_{0}^{\infty}\nu d\nu\right).$$
Here natural units are used. The paper then used the Euler-Maclaurin formula to get $\Delta{E} = -\frac{\pi}{12L}$
I read the Wikipedia on the formula and it looks like it is a method of evaluate the difference between closely related integrals and sums, but I don’t understand how we can use the formula for this scenario since we start our sum at $n=1$ and our integral at $n=0$. So my question is, how do we explicitly use the Euler-Maclaurin formula to get our desired result?
 A: 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.
A: I want to add something to @Andrews answer. They gave a good answer which is technically correct, but I think there should be at least some things added to Trick 3.
In the present answer there are some "mathematical non-rigorous" steps and the author claims that this is okay for physicists. The true reason why this is okay is that the expression,
\begin{equation}
\Delta E = \frac{\pi}{L} \left(\sum_{n=1}^\infty\nu -\int_1^\infty  \nu d \nu - \frac{1}{2}\right)\qquad(1)
\end{equation}
was wrong to begin with. And the reason that $$1 = \left[\lim_{\epsilon\rightarrow 0}e^{-\epsilon \nu}\right]$$
solves this problem is not only a mathematical trick but a physical argument which makes sense out of eq. (1). Let us remember what kind of system we are trying to describe. There are two conducting metal plates which result in boundary conditions for the electromagnetic field, because we know that the charges within the plate will move along the plate until the field satisfies the boundary conditions. But there is more physics than that! What about small wavelengths? Will the plates be able to shield electromagnetic waves of a shorter wavelength than let's say the inter-molecular distance within the plate? No, the photons of such a wavelength will just penetrate the metal plate, hence the boundary condition is violated for high-energy photons. This physical fact yields a UV-cutoff natural to the system! There is no mathematical non rigorous interchange of limits needed, but the initial formula was wrong physics-wise!
This goes even so far that the calculation is "independent" of the regulator scheme you choose to cut-off the short wavelength waves.
For a conceptual discussion of the matter at hand see e.g. "Quantum Field Theory and the Standard Model" by M.D. Schwartz.
