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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.

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  • $\begingroup$ If you want to add a remark about the mathematical meaning of the series expansion of an integral it is appreciated. $\endgroup$ – user243301 Feb 27 '17 at 22:17
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    $\begingroup$ Use the individual series expansions of both functions multiply them together and integrate term by term. $\endgroup$ – Zaid Alyafeai Feb 27 '17 at 22:18
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    $\begingroup$ 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 $\endgroup$ – reuns Feb 27 '17 at 22:21
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    $\begingroup$ You are welcome, you can use the Cauchy product formula to multiply both series. $\endgroup$ – Zaid Alyafeai Feb 27 '17 at 22:24
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    $\begingroup$ The Taylor series of $h(z ) = arctan(z)$ isn't complicated because $h'(z) = \frac{1}{1+z^2}$ ... $\endgroup$ – reuns Feb 27 '17 at 22:30
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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.

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  • $\begingroup$ Many thanks for these nice details. $\endgroup$ – user243301 Feb 28 '17 at 8:05
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    $\begingroup$ @user243301, I am not sure if the integral will have a nice closed form using that expansion. $\endgroup$ – Zaid Alyafeai Feb 28 '17 at 16:22
  • $\begingroup$ 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. $\endgroup$ – user243301 Feb 28 '17 at 18:28
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    $\begingroup$ @user243301, I checked the expansion using wolfram it is correct. Maybe you are considering the integral of the expansion I provided. $\endgroup$ – Zaid Alyafeai Feb 28 '17 at 18:46
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    $\begingroup$ @user243301 just multiply the series expansion I gave by xy and integrate. $\endgroup$ – Zaid Alyafeai Feb 28 '17 at 19:03

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