galactus wrote:

I have ran into this one before except the upper limit of integration was 1/2.

I have to go to work now, but if no one responds I believe I can write up a proof later today or tomorrow.

Thanks for your help in advance. I real need your help.

I want to clear the problem you noted above. In fact, whether the upper limit of the integration is 1 or 1/2 is allowed once you notice the fact that

\(\displaystyle \int_0^{1}B_{2n+1}(x)\cot({\pi}x)dx=2\int_0^{1/2}B_{2n+1}(x)\cot({\pi}x)dx\)

The equation also can be rewriten as

\(\displaystyle \int_0^{1/2}B_{2n+1}(x)\cot({\pi}x)dx=\int_{1/2}^{1}B_{2n+1}(x)\cot({\pi}x)dx\)

If we substitute \(\displaystyle x\) with \(\displaystyle 1-u\),then the right hand of the identity is(\(\displaystyle I\))

\(\displaystyle I=\int_{1/2}^{0}B_{2n+1}(1-u)\cot({\pi}(1-u))d(-u)=-\int_0^{1/2}B_{2n+1}(1-u)\cot({\pi}u)du\)

Utilizing the identity

\(\displaystyle B_{n}(1-x)=(-1)^{n}B_{n}(x)\)

we obtain

\(\displaystyle I=-\int_0^{1/2}(-1)^{2n+1}B_{2n+1}(u)\cot({\pi}u)du=\int_0^{1/2}B_{2n+1}(x)\cot({\pi}x)dx\)

I hope what I have explained can do some help to you, and I also suggest you to read the remarkable article below(maybe you have already read it):

http://www.sciencedirect.com/science/article/pii/S0377042702003588