Find a integral expression involving even degree Bernoulli polynomials that allows sum the series: $$P(k)=\sum_{n=1}^{\infty}\dfrac{\operatorname{Si}(n\pi)}{n^{2k+1}}$$ where Si(x) denotes the sine integral function defined by: $$\operatorname{Si}(x)=\int_0^x \dfrac{\sin(u)}{u}\mathrm{d}u$$
My work: We can express the Bernoulli Polynomial of degree m, B_m(x), as a Fourier series for $0\leq x<1$. If $m$ is odd, we have: $$ B_{2k+1}(x)=(-1)^{k-1}\dfrac{2(2k+1)!}{(2\pi)^{2k+1}}\sum_{n=1}^{\infty}\dfrac{\sin(2\pi n x)}{n^{2k+1}}\quad 0\leq x<1 $$ Then, we can divide by x and integrate from 0 to 1 both sides: $$ \int_0^1 \dfrac{B_{2k+1}(x)}{x}=C_k\int_0^1 \sum_{n=1}^{\infty}\dfrac{\sin(2\pi n x)}{x\cdot n^{2k+1}}\: dx=C_k\sum_{n=1}^{\infty}\dfrac{1}{n^{2k+1}}\int_0^1\dfrac{\sin(2\pi n x)}{x}\:dx=C_k\sum_{n=1}^{\infty}\dfrac{Si(2\pi n)}{n^{2k+1}} $$ But the last expression has a $\operatorname{Si}(2\pi n)$, and the statement of the problem is with $\operatorname{Si}(\pi n)$.
Just alter the integration slightly. The Bernoulli polynomial, $$ B_{2k+1}(x)=(-1)^{k-1}\dfrac{2(2k+1)!}{(2\pi)^{2k+1}}\sum_{n=1}^{\infty}\dfrac{\sin(2\pi n x)}{n^{2k+1}},\quad 0\leq x<1, $$ can be integrated to $x=+\tfrac{1}{2}$ rather than $1$, so that $$ \int_{0}^{1/2} \dfrac{B_{2k+1}(x)}{x}\,{\rm d}x= $$ $$ =C_{k}\int_{0}^{1/2} \sum_{n=1}^{\infty}\dfrac{\sin(2\pi n x)}{x\cdot n^{2k+1}}\: {\rm d}x=C_{k}\sum_{n=1}^{\infty}\dfrac{1}{n^{2k+1}}\int_{0}^{1/2}\dfrac{\sin(2\pi n x)}{2\pi n x}\:{\rm d}(2\pi n x), $$ $$=C_{k}\sum_{n=1}^{\infty}\dfrac{1}{n^{2k+1}}\int_{0}^{n\pi}\dfrac{\sin(u)}{u}\:{\rm d}u =C_{k}\sum_{n=1}^{\infty}\dfrac{\operatorname{Si}(n\pi)}{n^{2k+1}},$$ where $C_{k}=(-1)^{k-1}\dfrac{2(2k+1)!}{(2\pi)^{2k+1}}$, and in the third integral the numerator and denominator have both been multiplied by $2\pi n$ with the variable change $u=2\pi n x$. The last expression now has the required $\operatorname{Si}(n\pi )$, and you have expressed the series in terms of integrals of the Bernoulli polynomial from $x=0$ to $x=\tfrac{1}{2}$.