Limiting value of a contour integral Fix an integer $r \geq 1$, and a complex real $u$. (edit: I think $u$ is meant to be real)
For an integer $k \geq 1$ define
$$
a_k = \int_{ |z| = (2k+1)\pi } \frac{z \exp( u z)}{(e^z - 1)z^{2r+1}}dz.
$$
In the definition of $a_k$, the contour is meant in the counter-clockwise sense.
Then is there an easy way to prove $\lim_{k \to \infty} a_k = 0?$
Background : This is from an exercise in Tenenbaum's book on analytic number theory. The above leads to a way to determine the Fourier expansion of the even Bernoulli polynomials just using contour integration.
 A: Put $r_k=(2k+1)\pi$, $f(z)=\frac{e^{uz}}{z^{2r}(e^z-1)}$ and inscribe $\Gamma=\{z :|z|=r_k\}$ in a square $S$ so that $\Gamma$ can be continuously deformed into $S$. Using Cauchy's Thm. for homotopic curves gives,
$$\begin{align}\int_{\Gamma} f(z)\text{d}z&=\int_Sf(z)\text{d}z\\&=\int_{-r_k}^{r_k}f(t-ir_k)\text{d}t+\int_{-r_k}^{r_k}f(it+r_k)\text{d}t+\int_{-r_k}^{r_k}f(t+ir_k)\text{d}t+\int_{-r_k}^{r_k}f(r_k-it)\text{d}t \end{align} $$
Since the problem in Tenenbaum refers to the Bernoulli Polynomials $b_n(u)$, we assume $u\in[0,1]$. This implies $-1\le u-1 \le 0$, so $e^{(u-1)t}\le 1$ for $t\ge 0$ and $e^{(u-1)t}\le e^{-t}$ for $t<0$, hence $\frac{e^{ut}}{e^t+1}\le 1$ for all $t$. Also, if $1\le t$, $\frac12e^{(u-1)t}+e^{-t}\le 1$ so $e^{(u-1)t}\le2(1-e^{-t})\Rightarrow \frac {e^{ut}}{e^t-1}\le2$. Applying this in $(1),(3)$ along with the reverse triangle inequality in $(2)$ gives
$$\begin{align}\left|\int_{-r_k}^{r_k}f(t-ir_k)\text{d}t\right|&\le \int_{-r_k}^{r_k}\frac{e^{ut}}{|t-ir_k|^{2r}(e^t+1)}\text{d}t \overset{(1)}\le \int_{-r_k}^{r_k}\frac{1}{|t^2+r_k^2|^{r}}\text{d}t\le \frac 2{r_k^{2r-1}} \\ 
\left|\int_{-r_k}^{r_k}f(it+r_k)\text{d}t\right|&\overset{(2)}\le \int_{-r_k}^{r_k}\frac{e^{ur_k}}{|it+r_k|^{2r}(e^{r_k}-1)}\text{d}t \overset{(3)}\le 2\int_{-r_k}^{r_k}\frac{1}{|t^2+r_k^2|^{r}}\text{d}t\le \frac 4{r_k^{2r-1}}  \end{align} $$
The inequalities for the remaining two integrals follow similarly. Therefore, $\int_{\Gamma} f(z)\text{d}z\le O(r_k^{1-2r})\to 0$ as $k \to \infty$ for each $r\in \Bbb N^+$. 

Furthermore, since $e^z=(e^z-1)'$ has no roots, all nonzero poles of $f$ are simple. In particular, $\text{Res}(f,2\pi i m )=(2\pi im)^{-2r}e^{2\pi ium}$. At the origin, expand the generating function of the Bernoulli Polynomials, $g(z)=\frac{ze^{uz}}{e^z-1}=\sum_{n\ge 0}\frac{b_n(u)}{n!}z^n$ and use its uniform convergence on the disk $B(0,2\pi)$ to interchange summation and integration after applying the Residue Thm. thusly 
$$\begin{align}\int_{\Gamma} f(z)\text{d}z&= \sum_{m=-k}^k \text{Res}(f,2\pi i m) \\&=\frac 1{2\pi i}\sum_{n\ge 0}\frac{b_n(u)}{n!}\int_{|z|=1}z^{n-2r-1}\text{d}z+ \sum_{m=-k, m \neq 0}^k (2\pi i m)^{-2r}e^{2\pi i u m}\\ &= \frac{b_{2r}(u)}{(2r)!}+ \sum_{m=-k, m \neq 0}^k (2\pi i m)^{-2r}e^{2\pi i um}\end{align} $$
Now let $\Gamma$ expand to infinity and use the previous result to obtain:
$$\begin{align}b_{2r}(u)=-(2r)!\sum_{|m|\in \Bbb N^+} (2\pi i m)^{-2r}e^{2\pi i um} =\frac{(-1)^{r+1}(2r)!}{(2\pi)^{2r}}\sum_{|m|\in \Bbb N^+}   \frac{e^{2\pi i um}}{m^{-2r}} \end{align} $$
Since the LHS is real, the RHS is just the Cosine Series of $b_{2r}(u)$. Finally, use the fact that $\cos(-x)=\cos(x)$ to get:
$$b_{2r}(u)=\frac{(-1)^{r+1}(2r)!}{(2\pi)^{2r}}\sum_{|m|\in \Bbb N^+}   \frac{\cos(2\pi mu)}{m^{-2r}}=\frac{2(-1)^{r+1}(2r)!}{(2\pi)^{2r}}\sum_{m\ge1}   \frac{\cos(2\pi mu)}{m^{-2r}} $$
Putting $u=0$ yields
$$\zeta(2r)=(-1)^{r+1}\frac{B_{2r}(2\pi)^{2r}}{2(2r)!} $$
