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We wish to evaluate this integral, $$\int_{0}^{1}\arctan(x)\cdot\ln\frac{x+x^3}{(1-x)^2}\cdot\frac{\mathrm dx}{x}$$

I have tried using substitution, $u=\frac{x+x^3}{(1-x)^2}$ and integration by parts but, I am not able to simplify it down.

What can kind of method can we use to evaluate this integral?

Ali Shadhar
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Sibawayh
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    Wolfram|Alpha says that $\int_0^1\arctan x\ln x\frac{\mathrm dx}x=-\frac{\pi^3}{32}$, so that's a start, but it can't handle the other two factors. – joriki Dec 31 '19 at 07:27
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    Substituting $u=\frac1x$ yields

    \begin{eqnarray} \int_0^1\arctan x\ln\frac{1+x^2}{(1-x)^2}\frac{\mathrm dx}x &=& \int_1^\infty\arctan\frac1x\ln\frac{1+x^2}{(1-x)^2}\frac{\mathrm dx}x;. \ &=& \int_1^\infty\left(\frac\pi2-\arctan x\right)\ln\frac{1+x^2}{(1-x)^2}\frac{\mathrm dx}x;. \end{eqnarray}

    Unfortunately there's a minus sign, so this yields a way of obtaining the integral from $0$ to $\infty$ rather than a way to express the integral from $0$ to $1$ in terms of the integral from $0$ to $\infty$.

     – joriki Dec 31 '19 at 08:07
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    So

    $$ \int_0^\infty\arctan x\ln\frac{1+x^2}{(1-x)^2}\frac{\mathrm dx}x=\frac\pi2\int_1^\infty\ln\frac{1+x^2}{(1-x)^2}\frac{\mathrm dx}x;, $$

    which according to Wolfram|Alpha is $\frac{3\pi^3}{16}$. Not that this is any real progress; I'm just thinking out loud in case any of this happens to help.

     – joriki Dec 31 '19 at 08:17

1 Answers1

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$$I=\int_0^1\frac{\arctan x}{x}\ln\left(\frac{x+x^3}{(1-x)^2}\right)dx\\=\int_0^1\frac{\arctan x\ln x}{x}dx+\int_0^1\frac{\arctan x}{x}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx$$

For the first integral, write the taylor series of $\arctan x$

$$\int_0^1\frac{\arctan x\ln x}{x}dx=\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}\int_0^1x^{2n}\ln x\ dx=-\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^3}=-\beta(3)=-\frac{\pi^3}{32}$$

where $\beta(3)$ is the Dirichlet beta function.

as for the second one, it is evaluated here

$$\int_0^1\frac{\arctan x}{x}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx=\frac{\pi^3}{16}$$

$$\Longrightarrow I=\frac{\pi^3}{32}$$

Ali Shadhar
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    Why is the first integral trivial? – joriki Dec 31 '19 at 11:22
  • Just write the Taylor series of $\arctan x$ and see what you get. I can write the details if you like. – Ali Shadhar Dec 31 '19 at 15:00
  • Thanks. I can substitute the Taylor series and perform the integral, but I don't know how to evaluate the resulting sum $\sum_k(-1)^k/(2k+1)^3$. I found it evaluated in this answer, but I wouldn't exactly call that trivial :-) Is there an easier way to evaluate this sum? – joriki Dec 31 '19 at 15:25
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    What's left is the dirichlet beta function $\beta(3)$. – Ali Shadhar Dec 31 '19 at 15:28
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    https://en.m.wikipedia.org/wiki/Dirichlet_beta_function – Ali Shadhar Dec 31 '19 at 15:28
  • Aha, thanks a lot, good to know :-) – joriki Dec 31 '19 at 15:30
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    You welcome. $\beta(3)$ can be calculated differently but I didnt as it's already a well known value. – Ali Shadhar Dec 31 '19 at 15:35
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    @joriki more like is trivial in comparison to the other integral, since that one is quite difficult. For a way to find that sum, notice that: $$\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^3}=\sum_{n=1}^\infty \frac{\sin\left(\frac{n\pi} {2} \right)} {n^3}$$ And integrate twice the fourier series of $\frac{\pi-x} {2} $, similarly as seen here: https://math.stackexchange.com/a/3285129/515527 – Zacky Dec 31 '19 at 16:08
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    @Zacky: Thanks, that makes it more accessible to me :-) – joriki Dec 31 '19 at 16:31