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Can you help me find the primitive of $$\tan(x)\arctan(x)$$

Thanks in advance :)

Harry Peter
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Naoufal EL JAOUHARI
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  • What is the definition of primitive? – user25004 Jun 03 '14 at 18:54
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    @user25004 It's just a fancy name for the antiderivative, it is commonly used in French. – Hakim Jun 03 '14 at 18:55
  • Where did this problem come from? – Seth Jun 03 '14 at 18:57
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    @حكيمالفيلسوفالضائع It's not fancy at all. It's the classical term. And it has the advantage to be an actual word, unlike "antiderivative". – Daniel Fischer Jun 03 '14 at 18:57
  • @DanielFischer I used the term fancy because the OP used it here in this English site and that only a handful number of people would understand it. – Hakim Jun 03 '14 at 18:59
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    @حكيمالفيلسوفالضائع Sorry, I don't understand. It was until recently the term used in English. The neologism "antiderivative" is only a few decades old. – Daniel Fischer Jun 03 '14 at 19:01
  • I don't think it can be expressed in primitive functions... – m0nhawk Jun 03 '14 at 19:01
  • I don't think you can get the primitive of it. – Jika Jun 03 '14 at 19:01
  • I tried on Wolfram alpha and it timed out. That's why I was asking where this problem arose. – Seth Jun 03 '14 at 19:02
  • You always have recourse to the formula $F(x) = \int_0^x \tan(x) \arctan(x) , dx$. – Lee Mosher Jun 03 '14 at 19:05
  • I tried in Mathematica and in integrals.wolfram but no formula found, which strongly suggests that this integral may not be expressible in terms of elementary functions. – Hakim Jun 03 '14 at 19:09
  • I've been trying to find the exact value of $$ F(x) = \int_0^{\pi/2} \frac{1}{(1+\tan(x))(1+x^{2})}, dx $$, And "primitive" is French, I it's the classical term for antiderivative – Naoufal EL JAOUHARI Jun 03 '14 at 19:14
  • @Nel: Seriously? From where would you get that problem? :O

    It was discussed here before: http://math.stackexchange.com/questions/364452/how-to-evaluate-int-0-frac-pi2-frac11x21-tan-x-mathrm-d

     – Pranav Arora Jun 03 '14 at 19:23
  • @Daniel: Being an "actual word" (by which I presume you mean English word) is actually a disadvantage, because it carries with it the denotations and connotations from the English usage which may not be relevant (or even incredibly misleading) to the technical term. –  Jun 30 '14 at 14:31
  • @Hurkyl By "actual word", in this case, I meant "technical term long enough in use that I have learned it when I was young enough to actually learn things, and now get off my lawn you kids" ;) – Daniel Fischer Jun 30 '14 at 14:43

1 Answers1

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I am quite sure (and Mathematica supports my intuition) that there is no primitive in terms of standard functions. $\\ $ However, on any disk in the complex plane centred around the origin with radius < 1, each of the two factors is a holomorphic function and can be represented by their Taylor expansion. (c.f. Wikipedia, Taylor series for the individual coefficients and the radii of convergence). Because, within their radius of convergence, power series are absolutely convergent, the product of the two Taylor series is equal to their Cauchy product:

$\tan(x)\times \arctan(x) = \left( \sum\limits_{n=0}^\infty \underbrace{\frac{B_{2n}(-4)^n(1-4^n)}{(2n)!}}_{\equiv a_n} x^n\right)\times \left( \sum\limits_{m=0}^\infty \underbrace{\frac{(-1)^n}{2n+1}}_{\equiv b_m} x^m\right) =\sum\limits_{j=0}^\infty \left( \sum\limits_{k=0}^{\infty}a_k b_{j-k} \right)x^j$.

This can be integrated term by term (uniform convergence of power series) to yield:

$\int \tan(x)\times \arctan(x) dx= \sum\limits_{j=0}^\infty \frac{ \sum\limits_{k=0}^{\infty}a_k b_{j-k} }{j+1}x^{j+1}+const.$.

This is a series representation of the primitive that is valid for all $x, |x|<1$.

Cyclone
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