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I tried the following but it basically returns unevaluated:

In[1]:= D[ArcTan[x], {x, n}]
Out[1]= Derivative[n][ArcTan][x]

In[2]:= FunctionExpand[Derivative[n][ArcTan][x], Element[n, Integers] && n > 0]
Out[2]= Derivative[n][ArcTan][x]

Is it possible to request Mathematica to expand it as a function of two variables x, n?

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marked as duplicate by belisarius, Artes, Jens, Sjoerd C. de Vries, Silvia May 5 '13 at 22:51

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Have a look at this. –  b.gatessucks May 5 '13 at 17:11

1 Answer 1

up vote 13 down vote accepted

I know two approaches to this:

In[1]:= FullSimplify[SeriesCoefficient[ArcTan[y], {y, x, n}] n!, Element[n, Integers] && n > 0]
Out[1]= 1/2 I ((-I - x)^n - (I - x)^n) (1 + x^2)^-n Gamma[n]

and

In[2]:= FullSimplify[InverseFourierTransform[(-I k)^n FourierTransform[
          ArcTan[x], x, k] , k, x], Element[n, Integers] && n > 0]
Out[2]= 1/2 I (-(-1 - I x)^n + (1 - I x)^n) (I (1 + x^2))^-n Gamma[n]

The latter approach also could be used to find repeated indefinite integrals.

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