3
$\begingroup$

When I was studying $$\int_0^1\int_0^1\frac{x^\alpha y^\beta\arctan(xy)}{1-xy}dxdy,$$ (I tried get the case $\alpha=\beta$ and the general case with a Cauchy product) I've asked to Wolfram Alpha online calculator a simple case

integrate xy arctan(xy)/(1-xy)dx

and one of the outputs was the series expansion of the integral at $x=0$.

Question. Can you explain me how do you get the series expansion of the integral $$\int \frac{x\arctan(xy)}{1-xy}dx$$ at $x=0$? Thanks in advance.

  • 0
    If you want to add a remark about the mathematical meaning of the *series expansion of an integral* it is appreciated. – 2017-02-27
  • 1
    Use the individual series expansions of both functions multiply them together and integrate term by term. – 2017-02-27
  • 0
    Many thanks for your atention @ZaidAlyafeai I know you from this site for your nice integrals. – 2017-02-27
  • 1
    You mean the expansion of $f(x) = \int \frac{x\arctan(xy)}{1-xy}dx$ around $x=0$. You can deduce it easily from the expansion of $f'(x) =\frac{x\arctan(xy)}{1-xy}$. I upvoted because you didn't mention $\zeta$ but your question doesn't really deserve it – 2017-02-27
  • 1
    You are welcome, you can use the Cauchy product formula to multiply both series. – 2017-02-27
  • 0
    @user243301 Please also stop with all those useless comments. Do you understand what we wrote me and Zaid ? What is the Taylor series (in $x$) of $arctan(xy)$ and $1/(1-xy)$ ? can you multiply them together ... ? – 2017-02-27
  • 0
    Thanks @user1952009, I delete it, I read your comments and Zaid's comment. I know how multiply the series, then you say that is $f'$, but the output seems complicated, because it appear even the Catalan's constatn- – 2017-02-27
  • 1
    The Taylor series of $h(z ) = arctan(z)$ isn't complicated because $h'(z) = \frac{1}{1+z^2}$ ... – 2017-02-27
  • 0
    The series are $$\sum_{k=0}^\infty\frac{(-1)^k (xy)^{2k+1}}{2k+1}$$ and the geometric @user1952009 , where is the Catalan's constans? I will try study it. – 2017-02-27
  • 0
    ok so can you answer to your question now ? – 2017-02-27
  • 0
    The Cauchy product, that is convergent since $|x|<1$ and $|y|<1$ is an expansion series, maybe which that you want to write $\frac{\arctan(xy)}{1-xy}$, that is $$\sum_{n=0}^\infty\left(\sum_{k=0}^n\frac{(-1)^ky^{n+k+1}}{2k+1}x^{n+k+1}\right).$$ Any case, many thanks @user1952009 – 2017-02-27
  • 0
    This is not correct. Do you know how to multiply two power series ? – 2017-02-27
  • 0
    You are right now I see with the answer where was my mistake @user1952009 – 2017-02-28

1 Answers 1

2

Consider

$$\arctan(x) = \int^x_0 \frac{1}{1+t^2}\,dx = \sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{2k+1}$$

We can rewrite as

$$\sum_{k=0}^\infty a_k x^k$$

Where $a_{2j}=0$ and $a_{2j+1} = \frac{(-1)^j}{2j+1}$ , also we have

$$\frac{1}{1-x} = \sum_{k=0}^\infty b _kx^k$$

where $b_k = 1$

Then by Cauchy product formula we have

$$\left(\sum_{k=0}^\infty a_k x^k\right)\left(\sum_{k=0}^\infty b_k x^k\right) =\sum_{k=0}^\infty\left(\sum_{j=0}^k a_jb_{k-j}\right)x^k$$

We need to find

$$\sum_{j=0}^k a_jb_{k-j} = \sum_{j=0}^k a_j$$

Consider the two cases when $k =2n$ is even

$$ \sum_{j=0}^{2n} a_j = \sum_{j=0}^{n} a_{2j}+ \sum_{j=0}^{n-1} a_{2j+1} = \sum_{j=0}^{(k/2)-1} \frac{(-1)^j}{2j+1}$$

If $k = 2n+1$ is odd then

$$ \sum_{j=0}^{2n+1} a_j = \sum_{j=0}^{n-1} a_{2j}+ \sum_{j=0}^{n} a_{2j+1} = \sum_{j=0}^{(k-1)/2} \frac{(-1)^j}{2j+1}$$

Hence we know that

$$\sum_{j=0}^k a_j = \sum_{j=0}^{\lceil \frac{k}{2} \rceil -1} \frac{(-1)^j}{2j+1} $$

Finally we have

$$\frac{\arctan(xy)}{1-xy} = \sum_{k=0}^\infty \left(\sum_{j=0}^{\lceil \frac{k}{2} \rceil -1} \frac{(-1)^j}{2j+1} \right) (xy)^k $$

Note that the coefficients are variations of the alternating Harmonic numbers.

  • 0
    Many thanks for these nice details. – 2017-02-28
  • 1
    @user243301, I am not sure if the integral will have a nice closed form using that expansion. – 2017-02-28
  • 0
    Many thanks for your comment and answer, as you see several users vote your answer, I don't accept the answer because I am waiting to see why Wolfram Alpha provide a series expansion at $z=0$ of the integral as output that *seems* different of yours. Thus I am waiting feeback from the community. In few days I consider it. On the other hand I don't ask about the integral, you can add some remarks if you want as companion of your answer when you want. Thanks for your attention. – 2017-02-28
  • 1
    @user243301, I checked the expansion using wolfram it is correct. Maybe you are considering the integral of the expansion I provided. – 2017-02-28
  • 0
    What I would to know is also the output *Series expansion of the integral at x=0:* that shows Wolfram Alpha online calculator when you type the code: *integrate xy arctan(xy)/(1-xy)dx* – 2017-02-28
  • 1
    @user243301 just multiply the series expansion I gave by xy and integrate. – 2017-02-28
  • 0
    Then I'm sorry, many thanks for such details. – 2017-02-28
  • 0
    @user243301, no problem. Glad that helped you. – 2017-02-28