Euler numbers in terms of second kind Stirling numbers and in terms of the Touchard polynomials

Question

I recently managed to prove these equalities:

$$E_n=\sum_{k=0}^{n}S(n,k)\frac{-k!\sqrt{2}}{\sqrt{2}^{k}}\cos(\frac{3\pi}{4}(k+1))$$

$$E_n=2\int_{0}^{\infty}e^{-t}\cos(t)T_n(-t)dt$$

where $E_n$ are the Euler number, $S(n,k)$ the Stirling numbers of second kind and $T_n(x)$ the Touchard Polynomials.

My question is if they are useful, or if they are already discovered, since I didn't find anything.

For example, in Wikipedia we can see: (but I think this it's too much complicated). $$E_{n}=2^{2n-1}\sum_{\ell=1}^{n}\frac{(-1)^{\ell}S(n,\ell)}{\ell+1}\left(3\left(\frac{1}{4}\right)^{(\ell)}-\left(\frac{3}{4}\right)^{(\ell)}\right)$$

Answer 1

For the first one OP proposes

$$E_n = -\sqrt{2} \;\sum_{k=0}^n {n\brace k} \frac{k!}{\sqrt{2}^k} \cos(3\pi (k+1)/4).$$

which is

$$E_n = -\sqrt{2} \; \Re \sum_{k=0}^n {n\brace k} \frac{k!}{\sqrt{2}^k} \exp(3\pi i (k+1)/4).$$

Using the Stirling set number EGF this becomes

$$- \Re \left[ n! [z^n] \; (-1+i) \sum_{k=0}^n (\exp(z)-1)^k \frac{1}{\sqrt{2}^k} \exp(3\pi i k/4) \right].$$

We may extend the inner sum to infinity due to the coefficient extractor and the fact that $\exp(z)-1 = z + \cdots$, getting

$$\Re \left[ n! [z^n] \; (1-i) \frac{1}{1-\exp(3\pi i/4)(\exp(z)-1)/\sqrt{2}} \right] = \Re \; n! [z^n] f(z).$$

We have that $f(z)$ simplifies to

$$\frac{2(1-i)}{1+i+\exp(z)(1-i)} = \frac{2}{i+\exp(z)}.$$

For the real part we need the EGF of the conjugates. Using Mittag-Leffler we start with (poles of $f(z)$ are simple with residue $2i$)

$$g(z) = \sum_k \frac{2i}{z-(-\pi i/2 + 2\pi i k)} = \sum_k \frac{i(2z+\pi i)}{(z+\pi i/2)^2 - (2\pi i k)^2}$$

(the latter is convergent) and we can evaluate the sum through the residues in $w$ using the function

$$h(z,w) = - \frac{1}{w+(z+\pi i/2)} \frac{1}{w-(z+\pi i/2)} i(2z+\pi i) \cot(-iw/2) \frac{1}{2i}.$$

We use the fact that the residues in $w$ of $h(z,w)$ sum to zero and the residue at infinity is zero so that the residue at the two simple poles $w=\pm(z+ \pi i/2)$ produces for $g(z)$ (flip sign due to residue sum)

$$g(z) = - \frac{1}{2} \cot(-i(-(z+\pi i/2))/2) + \frac{1}{2} \cot(-i(+(z+\pi i/2))/2) \\ = - \frac{1}{2} \cot(iz/2-\pi/4) + \frac{1}{2} \cot(-iz/2+\pi/4) \\ = \cot(-iz/2+\pi/4).$$

We get for $g(z)$

$$\frac{\cos(-iz/2+\pi/4)}{\sin(-iz/2+\pi/4)} \\ = i \frac{\exp(z/2)\exp(\pi i/4)+\exp(-z/2)\exp(-\pi i/4)} {\exp(z/2)\exp(\pi i/4)-\exp(-z/2)\exp(-\pi i/4)} \\ = i \frac{\exp(z)\exp(\pi i/4)+\exp(-\pi i/4)} {\exp(z)\exp(\pi i/4)-\exp(-\pi i/4)} = i\frac{\exp(z)i+1}{\exp(z)i-1} = \frac{\exp(z)i+1}{i+\exp(z)}.$$

The difference between $f(z)$ and $g(z)$ is exactly $-i$ and we finally have

$$f(z) = -i + \sum_k \frac{2i}{z-(-\pi i/2 + 2\pi i k)}.$$

We seek the generating function of the conjugates which can now be obtained by inspection and is seen to be (expand terms into a series about zero)

$$f_C(z) = i + \sum_k \frac{-2i}{z-(\pi i/2 - 2\pi i k)} = i - \sum_k \frac{2i}{z-(\pi i/2 - 2\pi i k)}.$$

Here we have applied conjugation to

$$\frac{C}{z-\rho} = -\frac{C}{\rho} \frac{1}{1-z/\rho} = -\frac{C}{\rho} \sum_{q\ge 0} \frac{z^q}{\rho^q}.$$

We claim this is $-f(z-\pi i)$ and check

$$- f(z-\pi i) = i - \sum_k \frac{2i}{z-\pi i -(-\pi i/2 + 2\pi i k)} = i - \sum_k \frac{2i}{z-(\pi i/2 + 2\pi i k)}.$$

We are iterating over $k$ in two possible directions. Returning to the original problem we have that our answer is given by

$$\Re \; n! [z^n] f(z) = \frac{1}{2} n! [z^n] (f(z)+f_C(z)).$$

Using the expression for $f_C(z)$ in terms of $f(z)$ this becomes

$$\frac{1}{i+\exp(z)} - \frac{1}{i+\exp(z-\pi i)} = \frac{1}{i+\exp(z)} - \frac{1}{i-\exp(z)} \\ = \frac{i-\exp(z)-i-\exp(z)}{(-1)-\exp(2z)} = \frac{2\exp(z)}{\exp(2z)+1} = \frac{1}{\cosh(z)}.$$

We have obtained the EGF of the Euler numbers and may conclude.

Remark. The computation of the residue at infinity being zero may be seen at the following MSE link.

Answer 2

For the second one OP proposes in terms of Touchard polynomials

$$E_n = 2 \int_0^\infty \exp(-t) \cos(t) T_n(-t) \; dt.$$

This is

$$2 \sum_{k=0}^n {n\brace k} \; \Re \; \int_0^\infty \exp(-t(1-i)) (-1)^k t^k \; dt.$$

Now put $(1-i)t = u$ so that $\frac{1}{2} (1+i) u = t$ and we get

$$\sum_{k=0}^n {n\brace k} \frac{1}{2^k} (-1)^k \; \Re \; (1+i)^{k+1} \int_0^{\infty (1-i)} \exp(-u) u^k \; du.$$

Evaluating the integral with the Gamma function we get

$$\sum_{k=0}^n {n\brace k} \frac{1}{2^k} (-1)^k \; \Re \; (1+i)^{k+1} k!.$$

This has EGF

$$n! [z^n] \Re \left[ (1+i) \sum_{k\ge 0} (\exp(z)-1)^k \frac{1}{2^k} (-1)^k (1+i)^k \right]$$

where we have extended to infinity due to $\exp(z)-1=z+\cdots$. This is

$$n! [z^n] \;\Re\; \frac{1+i}{1+(\exp(z)-1)(1+i)/2} = n! [z^n] \;\Re\; f(z).$$

We have that $f(z)$ simplifies to

$$\frac{2(1+i)}{1-i+(1+i)\exp(z)} = \frac{2}{-i + \exp(z)} = - \frac{2}{i-\exp(z)}.$$

Note that we learned in the companion answer that the EGF of the conjugates of $f(z)$ is $\frac{2}{i+\exp(z)}.$ We thus obtain one more time that

$$\frac{1}{2} (f(z) + f_C(z)) = \frac{1}{\cosh(z)}$$

and may conclude.

Remark. For the Gamma function evaluation we use a pizza slice contour with an angle of $-\pi/4$ which contains no poles. So to apply the Gamma function we just need to show that the contribution from the arc $Q$ vanishes in the limit.

We get with $z=R\exp(i\theta)$ and $dz = iR\exp(i\theta) \; d\theta$

$$\int_Q \exp(-z) z^k \; dz = \int_{-\pi/4}^0 \exp(-R \exp(i\theta)) R^k \exp(ki\theta) iR\exp(i\theta) \; d\theta.$$

We have as an upper bound on the norm of this integral

$$\int_{-\pi/4}^0 \exp(-R\cos(\theta)) R^{k+1} \; d\theta \lt R^{k+1} \exp(-R/\sqrt{2}) \int_{-\pi/4}^0 1 \; d\theta \\ = \frac{\pi}{4} R^{k+1} \exp(-R/\sqrt{2}) \rightarrow 0 $$

as $R\to\infty.$