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Well I found the answer to your question, it is

$$\sum_x \tan(x)=ix-\psi _{e^{2 i}}^{(0)}\left(x+\frac{\pi }{2}\right)+C$$

I have verified it with difference operator and it gives tan(x). The function involved is the q-digamma function http://mathworld.wolfram.com/q-PolygammaFunction.html .

You can verify the result yourself: http://tiny.cc/60mmf

Well I found the answer to your question, it is

$$\sum_x \tan(x)=ix-\psi _{e^{2 i}}^{(0)}\left(x+\frac{\pi }{2}\right)+C$$

I have verified it with difference operator and it gives tan(x). The function involved is the q-digamma function http://mathworld.wolfram.com/q-PolygammaFunction.html .

You can verify the result yourself.

Well I found the answer to your question, it is

$$\sum_x \tan(x)=ix-\psi _{e^{2 i}}^{(0)}\left(x+\frac{\pi }{2}\right)+C$$

I have verified it with difference operator and it gives tan(x). The function involved is the q-digamma function http://mathworld.wolfram.com/q-PolygammaFunction.html .

You can verify the result yourself:

 http://tiny.cc/60mmf

Well I found the answer to your question, it is

$$\sum_x \tan(x)=ix-\psi _{e^{2 i}}^{(0)}\left(x+\frac{\pi }{2}\right)+C$$

I have verified it with difference operator and it gives tan(x). The function involved is the q-digamma function http://mathworld.wolfram.com/q-PolygammaFunction.html .

You can verify the result yourself:  http://tiny.cc/60mmf

Well I found the answer to your question, it is

$$\sum_x \tan(x)=ix-\psi _{e^{2 i}}^{(0)}\left(x+\frac{\pi }{2}\right)+C$$

I have verified it with difference operator and it gives tan(x). The function involved is the q-digamma function http://mathworld.wolfram.com/q-PolygammaFunction.html .

You can verify the result yourself:

http://www.wolframalpha.com/input/?i=DifferenceDelta(i+x-QPolyGamma(Pi%2F2%2Bx%2CE^(2+I))%2Cx)

Well I found the answer to your question, it is

$$\sum_x \tan(x)=ix-\psi _{e^{2 i}}^{(0)}\left(x+\frac{\pi }{2}\right)+C$$

I have verified it with difference operator and it gives tan(x). The function involved is the q-digamma function http://mathworld.wolfram.com/q-PolygammaFunction.html .

You can verify the result yourself:

http://tiny.cc/60mmf