3

Prove that

$$\int_0^1\arctan x\left(\frac{3\ln(1+x^2)}{1+x}-\frac{2\ln(1+x)}{x}\right)\ dx=\frac{3\pi}{8}\ln^22-\frac{\pi^3}{32}$$

I managed to prove the above equality using integral manipulation (solution be posted soon) , but is it possible to do it in different ways and specifically by harmonic series?

The interesting thing about this problem is that we don't see any imaginary part which is usually involved in such integrals.

Note: the second integral $$\int_0^1 \frac{\arctan x\ln(1+x)}{x}\ dx=\frac{3\pi^3}{32}+\frac{3\pi}{16}\ln^22+\frac32G\ln2+3\text{Im}\operatorname{Li}_3(1-i)$$ was already evaluated here but I calculated the original problem without separating the two integrals.

Ali Shadhar
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  • I'm looking forward to your solution. –  Jul 03 '19 at 18:21
  • I will provide soon. as i mentioned i got lucky with this problem. some nice cancellation and symmetry was there. – Ali Shadhar Jul 03 '19 at 18:25
  • I guess there are a bunch of similar integrals like this, that when substracted produces a nice result (the ugly terms are canceled). See here another one: https://math.stackexchange.com/q/3054741/515527 – Zacky Jul 03 '19 at 18:37
  • Anyway, your integral would look much better in the following form: $$\int_0^1 \frac{\arctan x\ln(1+x^2)}{x(1+x)}dx=\frac{\pi^3}{96}-\frac{\pi}{8}\ln^2 2$$ – Zacky Jul 03 '19 at 18:38
  • @Zacky thank you but my solution leads to the form of the problem. but based for the form you suggested, what happened to the other integral that has $\ln(1+x)/x$? – Ali Shadhar Jul 03 '19 at 18:44
  • @AliShather Well, just use this: https://math.stackexchange.com/questions/3006106/an-amm-like-integral-int-01-frac-arctan-xx-ln-frac1x231x2dx/3007623#3007623 $${\int_0^1 \frac{\arctan x \ln(1+x)}{x}dx}=\frac{3}{2}\int_0^1 \frac{\arctan x\ln(1+x^2)}{x}dx$$ The $2$ cancels out nicely, and $3$ is a common factor. – Zacky Jul 03 '19 at 19:02
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    @Zacky got you now. yes thats a nice equality i remember it very well. we can related these two equalities in different post. – Ali Shadhar Jul 03 '19 at 19:13
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    @Zacky yes Zacky thats what i did. I even copied the first two lines from you. too lazy to write :) – Ali Shadhar Jul 03 '19 at 19:14
  • Such integral are not that hard to get heuristically with the help of lindep function of PARI GP. – FDP Jul 04 '19 at 21:10
  • @FDP I never used such software. all I use is Wolfram and just to verify my solutions and basic indefinite integrals. – Ali Shadhar Jul 04 '19 at 21:24
  • I have found out the following example: \begin{align}\int_0^1\frac{\ln(1+x)}{1+x^2}\left(\frac{3\arctan x}{x}+2\ln x\right)dx=\frac{5}{128}\pi^3-\frac{7}{4}\text{G}\ln 2+\frac{3}{16}\pi \ln^2 2\end{align}I don't know how to prove this right now. – FDP Jul 04 '19 at 21:34
  • @FDP beautiful... I will work on it today. can you post it as a problem? – Ali Shadhar Jul 04 '19 at 21:36
  • Not sure my question will be accepted. – FDP Jul 04 '19 at 21:39

1 Answers1

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@Kemono Chen elegantly proved here $$\int_0^y\frac{\ln(1+yx)}{1+x^2}dx=\frac12 \tan^{-1}(y)\ln(1+y^2)$$

By integration by parts we have

$$\int_0^y\frac{y\tan^{-1}(x)}{1+yx}dx=\frac12 \tan^{-1}(y)\ln(1+y^2)$$ divide both sides by $1+y$ and integrate from $y=0$ to $y=1$, we get, \begin{align} \frac12I&=\frac12\int_0^1\frac{\arctan y\ln(1+y^2)}{1+y}\ dy=\int_0^1\int_0^y\frac{y\arctan x}{(1-yx)(1+y)}\ dy\ dx\\ &=\int_0^1\arctan x\left(\int_x^1\frac{y}{(1-yx)(1+y)}\ dy\right)\ dx\\ &=\int_0^1\arctan x\left(\frac{\ln(1+x)}{x}-\frac{\ln(1+x^2)}{x}+\frac{2\ln(1+x)-\ln(1+x^2)-\ln2}{1-x}\right)\ dx\\ &=\underbrace{\int_0^1\frac{\arctan x\ln(1+x)}{x}\ dx}_{J}-\underbrace{\int_0^1\frac{\arctan x\ln(1+x^2)}{x}\ dx}_{K}\\ &\quad+\underbrace{\int_0^1\frac{\arctan x}{1-x}\left(2\ln(1+x)-\ln(1+x^2)-\ln2\right)\ dx}_{\large{x=(1-y)/(1+y)}}\\ &=J-K-\int_0^1\frac{\left(\frac{\pi}{4}-\arctan x\right)}{x(1+x)}\ln(1+x^2)\ dx\\ &=J-K-\frac{\pi}{4}\int_0^1\frac{\ln(1+x^2)}{x(1+x)}\ dx+K-I\\ \frac32I&=J-\frac{\pi}{4}\left(\frac{3\pi^2}{48}-\frac34\ln^22\right)\\ 3I-2J&=\frac{3\pi}{8}\ln^22-\frac{\pi^3}{32} \end{align}

Ali Shadhar
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