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I have an expression which is simply

(j/k) x^(j/k) LerchPhi[x,1,j/k)] 

where 0 < j < k.

Manually I have been able (tedious work) to obtain only ArcTan's and ArcTanh's. How could I ask Mathematica to do this automatically in the most compact form ?

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Clarify your question. What exactly do you intend to do with that expression? Simplify or what? –  RunnyKine Oct 26 '13 at 9:00
    
@RunnyKine, I think the OP is looking for a generalisation of this. –  Simon Woods Oct 26 '13 at 9:05
    
@SimonWoods. Ah I see. –  RunnyKine Oct 26 '13 at 9:12
    
@RunnyKine. It is a kind of simplification. I have been able to do this by hand and arrived to quite nice linear combinations of ArcTan's and ArcTanh's. What I would like to learn is how to formulate the problem and get the most compact form of the result (ideally, one ArcTan and one ArcTanh, but linear combinations will be good). I am not very good (this is an understatement). –  Claude Leibovici Oct 26 '13 at 9:18
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1 Answer

This should get you started. For some reason Mathematica returns LerchPhi[x, 1, j/k] unevaluated if the third argument is a symbol. So let's assume $j = 1$ and $k = 4$ (it does meet your requirement $0 < j < k$ ).

Let's first define some complexity function:

cf[k_][e_] :=  k Count[e, _Log | _ArcCot | _ArcCoth, Infinity] + LeafCount[e]

Now we simplify:

FullSimplify[ FunctionExpand[(1/4) x^(1/4) LerchPhi[x, 1, 1/4]], 
              ComplexityFunction -> cf[#]] & /@ Range[4]
{1/2 (ArcTan[x^(1/4)] + ArcTanh[x^(1/4)]), 1/2 (ArcTan[x^(1/4)] + ArcTanh[x^(1/4)]),
 1/2 (ArcTan[x^(1/4)] + ArcTanh[x^(1/4)]),   1/2 (ArcTan[x^(1/4)] + ArcTanh[x^(1/4)])}

Looks like it converged fast to a nicely simplified expression in terms of only ArcTan and ArcTanh

1/2 (ArcTan[x^(1/4)] + ArcTanh[x^(1/4)])

Admittedly, for this lone case, FunctionExpand will do just fine, but if you're trying to generalize this, you'll need to use the ComplexityFunction. To see this try:

FunctionExpand[(1/8) x^(1/8) LerchPhi[x, 1, 1/8]]

You'll get your answer in terms of Log only

x^(1/8) ( -(Log[1 - x^(1/8)]/(8 x^(1/8))) 
          + (I Log[1 - I x^(1/8)])/(8 x^(1/8)) 
          - (I Log[1 + I x^(1/8)])/(8 x^(1/8)) 
          + Log[1 + x^(1/8)]/(8 x^(1/8)) 
          - ((-1)^(1/4) Log[1 - E^(-((I π)/4)) x^(1/8)])/(8 x^(1/8)) 
          + ((-1)^(3/4) Log[1 - E^((I π)/4) x^(1/8)])/(8 x^(1/8)) 
          - ((-1)^(3/4) Log[1 - E^(-((3 I π)/4)) x^(1/8)])/(8 x^(1/8)) 
          + ((-1)^(1/4) Log[1 - E^((3 I π)/4) x^(1/8)])/(8 x^(1/8)))

But if we apply our cf:

FullSimplify[FunctionExpand[(1/8) x^(1/8) LerchPhi[x, 1, 1/8]], ComplexityFunction -> cf[1]]

We obtain:

1/4 (ArcTan[x^(1/8)] + ArcTanh[x^(1/8)] - (-1)^( 3/4) (ArcTan[(-1)^(1/4) x^(1/8)] 
+ ArcTanh[(-1)^(1/4) x^(1/8)]))

Again, in terms of only ArcTan and ArcTanh.

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This looks as magics ! It is almost what I am looking for.If I may ask : is it feasible to remove the roots of (-1) and use automatically the multiple angles formulas for ArcTan and ArcTanh ? This is what I made by hand. Thanks for you help to the old idiot. –  Claude Leibovici Oct 26 '13 at 11:48
    
It does not seem to like (1/3) –  Claude Leibovici Oct 26 '13 at 11:55
    
@ClaudeLeibovici. Hmmm, interesting. I wonder why. Looking into it. –  RunnyKine Oct 26 '13 at 12:02
    
I do not know how to post a Word document to show you my formulas (if you are interested). In any manner, many thanks for your help from the old idiot ! –  Claude Leibovici Oct 26 '13 at 13:41
    
@ClaudeLeibovici. I'm glad I could help. If you have a cloud storage you could upload it there and put a link to it here. –  RunnyKine Oct 26 '13 at 13:49
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