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When given the following integral,

Integrate[HurwitzZeta[Log[2, u], q], u]

Mathematica returns it unevaluated. The answer should be

HurwitzZeta[Log[2, u], q/2] - u HurwitzZeta[Log[2, u], q] + HurwitzZeta[Log[2, u], (q + 1)/2] + C[1]

What should I change so that Mathematica will accept such integrals? Is there another method I could use?

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How did you get this answer from Mathematica? which version are you using? It remains unevaluated in V9. May be you meant Integrate[HurwitzZeta[Log[2, u], q], q] instead of Integrate[HurwitzZeta[Log[2, u], q], u] – Nasser Feb 9 at 21:08
@Nasser The OP hasn't said that he got the answer from Mma ... – belisarius Feb 9 at 21:10
1  
@belisarius, I see. But it looks like it was cut/paste from Mathematica the way it is written. May be OP should have made it more clear by saying the answer SHOULD come out as or such. – Nasser Feb 9 at 21:12
10  
The result you're trying to get isn't correct, as can be verified by setting q = 2 and taking the derivative of your integral with respect to u. Evaluate the derivative at u = 10. to see that the answer is significantly different from HurwitzZeta[Log[2, u], 2]. – Jens Feb 9 at 21:58
On version 9, setting q to a number (e.g. 2) gives \[Integral](-1 + Zeta[Log[u]/Log[2]]) \[DifferentialD]u which isn't quite "returning unevaluated". I'm also a bit puzzled why you'd expect an analytical solution for any q when, for at least some values of q<0, there are non-zero imaginary parts in your integrand. – Verbeia♦ Feb 11 at 3:33
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closed as too localized by Szabolcs, Sjoerd C. de Vries, Verbeia♦ Feb 13 at 22:25

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