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Earlier, I posted the sum evaluation that has the following integral representation.

$$\sum _{k=1}^{\infty }\frac{\left(-1\right)^kH_k}{\left(2k+1\right)^2}=\int _0^1\frac{\ln \left(x\right)\ln \left(1+x^2\right)}{1+x^2}\:\mathrm{d}x$$ Ali Shather managed to prove here $$\int _0^1\frac{\ln \left(x\right)\ln \left(1+x^2\right)}{1+x^2}\:\mathrm{d}x=\frac3{32}\pi^3+\frac{\pi}8\ln^22-\ln2~G-2\text{Im}\operatorname{Li_3}(1+i)$$ which relies on trigonometric substitutions as well as complex methods.

Question: Can this integral be evaluated without the trig functions? and is it possible to evaluate it without complex methods?

I tried using certain substitutions, but ended up with similar or even harder integrals. I'm not sure how to approach this.

Quanto
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  • Do you allow power series methods? – Eric Towers Aug 14 '20 at 20:48
  • Sure, as long your methods dont make use of trigonometric functions and if possible avoid complex methods. –  Aug 14 '20 at 21:00
  • I thought I was doing something clever with power series, but just ended up getting the original sum. Time to try something else now. – Varun Vejalla Aug 14 '20 at 21:34
  • Trying to avoid complex methods seems almost impossible, as the last term $\text{Im } \text{Li}_3(1+i)$ cannot be simplified anymore? – Tom Chen Aug 15 '20 at 05:15
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    @TomChen i agree, at some point complex methods are involved but it would be nice to see an approach that doesnt necessarily depend on them, let's say i manage to end up with an expression having $\displaystyle \int _0^1\frac{\ln ^2\left(1-x\right)}{1+x^2}:dx:\text{or}::\int _0^1\frac{\ln ^2\left(1+x\right)}{1+x^2}:dx$ these have to be done using complex methods but have been evaluated before in easy ways in this site. –  Aug 15 '20 at 06:18
  • @TomChen So I'm allowing to use the values of these kind of minor integrals is what I'm trying to say. –  Aug 17 '20 at 03:19
  • One can show that $\int_0^1 \frac{\ln x\ln\left(\frac{1+x^2}{2(1-x)^2}\right)}{1+x^2}dx=0$, then extract the desired integral out as $\int_0^1 \frac{\ln x\ln(1-x)}{1+x^2}dx$ can be found here. – Zacky Aug 20 '20 at 13:49
  • Without trigonometric functions? The $\pi$ has to come from somewhere. – J.G. Oct 19 '20 at 14:15

2 Answers2

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A real-method based solution. Integrate

\begin{align} &\int_0^1\frac{\ln x \ln (1+x^2)}{1+x^2}dx \\ =& \int _0^1 \ln (1+x^2)\> d \left(\int_1^x \frac{\ln t}{1+t^2}dt\right)\\ \overset{ibp}=&-\int _0^1 \frac{2x}{1+x^2}\left( \int_0^x \right. {\left. \frac{\ln t}{1+t^2} \overset{t=xs } {dt}+G\right)}dx \\ =& -\int_0^1 \int_0^1 \frac{2x^2\ln (xs) }{(1+x^2)(1+x^2s^2)}dsdx -G\int_0^1 \frac{2x}{1+x^2}dx \\ = &-\int_0^1 \int_0^1 \frac{2x^2\ln x}{(1+x^2)(1+x^2s^2)}dsdx \\ &\>\>\> - \int_0^1 \int_0^1 \frac{2x^2\ln s }{(1+x^2)(1+x^2s^2)}dsdx -G\ln2 \\ = &-2\int_0^1 \frac{x\ln x\> \tan^{-1}x}{1+x^2}dx -2\int_0^1 \frac{s\ln s\> \tan^{-1}s}{1-s^2}ds -G\ln2 \\ =& -4\int_0^1 \underset{=J}{\frac{x\ln x\> \tan^{-1}x}{1-x^4}}dx -G\ln2=-4J -G\ln2 \\ &\int _0^1\frac{\ln x \ln (1-x^2)}{1+x^2}dx \\ = &\int _0^1 \ln (1-x^2)\> d \left(\int_1^x \frac{\ln t}{1+t^2}dt\right) = 4J-2G\ln2 +\frac{\pi^3}{16}\\ \end{align}

Then, eliminate the common term $J$ $$\int_0^1\frac{\ln x\ln \left(1+x^2\right)}{1+x^2}dx = -\int _0^1\frac{\ln x \ln (1-x^2)}{1+x^2}dx -3G\ln 2 +\frac{\pi^3}{16}\tag1 $$

Next, combine \begin{align} \int_0^\infty \frac{\ln^2(1+x)}{1+x^2}& dx \overset{\text{split}(0,\infty)}= 2\int_0^1 \frac{\ln^2(1+x)}{1+x^2}dx\\ & -2 \int_0^1 \frac{\ln x\ln(1+x)}{1+x^2}dx+\frac{\pi^3}{16}\\ \int_0^1 \frac{\ln^2(1-x)}{1+x^2}& \overset{x\to\frac{1-x}{1+x}} {dx} = \int_0^1 \frac{\ln^2(1+x)}{1+x^2}dx\\ & -2 \int_0^1 \frac{\ln x\ln(1+x)}{1+x^2}dx-2G\ln2+\frac{\pi^3}{16}\\ \end{align} to get $$\int_0^1 \frac{\ln x\ln(1+x)}{1+x^2}dx =\frac12\int_0^\infty \frac{\ln^2(1+x)}{1+x^2}dx-\int_0^1 \frac{\ln^2(1-x)}{1+x^2}dx \\ -2G\ln2+\frac{\pi^3}{32}\tag2 $$ Also $$\hspace{-5mm}\int_0^1 \frac{\ln x\ln(1-x)}{1+x^2}dx =\frac12\int_0^1\frac{\ln^2(1-x)}{1+x^2}dx-\frac12\int_0^1\frac{\ln^2\frac x{1-x}}{1+x^2}dx + \frac{\pi^3}{32} \tag3 $$ Then, (2) + (3) $$ \int_0^1 \frac{\ln x\ln(1-x^2)}{1+x^2}dx =-\int_0^1 \frac{\ln^2(1-x)}{1+x^2}dx -2G\ln2+\frac{\pi^3}{16} +\frac12 K \tag4$$ where $ \int_0^1 \frac{\ln^2(1-x)}{1+x^2}dx=2\text{Im}\>\text{Li}_3\left(\frac{1+i}2\right)$ and \begin{align} K =& \int_0^1 \underset{t=1-x}{\frac{\ln^2(1-x)}{1+x^2}}dx +\underset{t=1+x}{ \int_0^\infty \frac{\ln^2(1+x)}{1+x^2}}dx -\int_0^1\underset{t=\frac x{1-x}}{\frac{\ln^2\frac x{1-x}}{1+x^2}}dx \\ =& \int_0^\infty {\frac{4t \ln^2 t}{4+t^4}}dt \overset{t^2\to 2t}= \frac{\pi}{8}\ln^22+\frac{\pi^3}{32} \end{align}

Plug into (4) and then into (1) to obtain $$ \int_0^1 \frac{\ln x\ln(1+x^2)}{1+x^2}dx =2\text{Im}\>\text{Li}_3\left(\frac{1+i}2\right) -G\ln2 -\frac{\pi}{16}\ln^22-\frac{\pi^3}{64}$$

Quanto
  • 120,125
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    The harder part is to prove that $\displaystyle \int_0^1 \frac{\ln^2(1-x)}{1+x^2}dx=2\text{Im}>\text{Li}_3\left(\frac{1+i}2\right)$ – FDP Oct 26 '20 at 15:36
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    It follows directly from the integral definition of $\text{Li}_3(a)$ $$\text{Li}_3(a)= \frac 12\int_0^1 \frac{a\ln^2t}{1-at}dt = \frac 12 \int_0^1 \frac{a\ln^2 (1-x)}{1-a+ax}dx$$ – Quanto Jan 12 '21 at 18:10
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Let us generalize the given integral

$$I(a,n)=\int _0^1\frac{x^a\ln^n \left(x\right)\ln \left(1+x^2\right)}{1+x^2}\:\mathrm{d}x$$

All we need is

$$\ln^2\left(1+x^2\right)=2\sum _{k=2}^{\infty } \frac{\left(-1\right)^kH_{k-1}}{k}x^{2k}$$ Let us now differentiate this expression with respect to $x$:

$$\frac{\ln \left(1+x^2\right)}{1+x^2}=\sum _{k=1}^{\infty }\left(-1\right)^{k-1}H_{k}x^{2k}$$ Next, we calculate the integral

$$I(a)=\int _0^1\frac{x^a\ln \left(1+x^2\right)}{1+x^2}\:\mathrm{d}x=$$

$$= \sum _{k=1}^{\infty }\frac{\left(-1\right)^{k-1}H_{k}}{2k+1+a}$$

The last step is to differentiate the obtained result by $a$ $n$ times:

$$I(a,n)=\left(-1\right)^nn!\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k-1}H_{k}}{(2k+1+a)^{n+1}}$$

Martin Gales
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