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I would like some help to find a closed form for the following integral:$$\int_0^1 \frac{\left(\log (1+x)\right)^3}{x}dx $$ I was told it could be calculated in a closed form. I've already proved that $$\int_0^1 \frac{\log (1+x)}{x}dx = \frac{\pi^2}{12}$$ using power series expansion.

Thank you.

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  • $\begingroup$ With the help of Mathematica; The closed form includes a polylogarithm and the riemann zeta function(Apery's constant to be exact). $\endgroup$ – UserX Oct 9 '14 at 17:56
  • $\begingroup$ @UserX Right, W/A gives a closed form too! $\endgroup$ – G. Cardano Oct 9 '14 at 17:58
  • $\begingroup$ W|A uses Mathematica. $\endgroup$ – UserX Oct 9 '14 at 18:07
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Letting $u = \log(1+x)$,

$$ \begin{align} \int_{0}^{1} \frac{\log^{3}(1+x)}{x} \ dx &= \int_{0}^{\log 2} \frac{u^{3}}{e^{u}-1} e^{u} \ du \\ &= \int_{0}^{\log 2} \frac{u^{3}}{1-e^{-u}} \ du \\ &= \int_{0}^{\log 2} u^{3} \sum_{n=0}^{\infty} e^{-nu} \ du \\ &= \sum_{n=0}^{\infty} \int_{0}^{\log 2} u^{3} e^{-nu} \ du \\ &= \int_{0}^{\log 2} u^{3} \ du + \sum_{n=1}^{\infty}\int_{0}^{\log 2} u^{3} e^{-nu} \ du \\ &= \frac{\log^{4}(2)}{4} + \sum_{n=1}^{\infty}\int_{0}^{\log 2} u^{3} e^{-nu} \ du .\end{align}$$

Then integrating by parts 3 times,

$$\begin{align} \int_{0}^{1} \frac{\log^{3}(1+x)}{x} \ dx &= \frac{\log^{4}(2)}{4} -\sum_{n=1}^{\infty} e^{-nu} \left(\frac{6}{n^{4}} + \frac{6u}{n^{3}} + \frac{3 u^{2}}{n^{2}} + \frac{u^{3}}{n} \right)\Bigg|^{\log 2}_{0} \\ &= \frac{\log^{4}(2)}{4} - \sum_{n=1}^{\infty} \left[\frac{1}{2^{n}} \left(\frac{6}{n^{4}} + \frac{6 \log 2}{n^{3}} + \frac{3 \log^{2} (2)}{n^{2}} + \frac{\log^{3}(2)}{n}\right) - 6 \zeta(4) \right] \\ &= -\frac{3\log^{4}(2)}{4 } - 6 \text{Li}_{4} \left(\frac{1}{2} \right) - 6 \log (2) \ \text{Li}_{3} \left(\frac{1}{2} \right)-3 \log^{2}(2) \text{Li}_{2} \left(\frac{1}{2} \right) + 6 \zeta(4) \\ &\approx 0.1425141979 . \end{align}$$

The answer could of course be simplified using the known values of $\text{Li}_{2} \left(\frac{1}{2} \right)$, $\text{Li}_{3} \left( \frac{1}{2}\right) $, and $\zeta(4)$.

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  • $\begingroup$ @JackD'Aurizio Thanks. $\endgroup$ – Random Variable Oct 9 '14 at 22:06
  • $\begingroup$ Random Variable, very clear to me now! thank you! +1. $\endgroup$ – G. Cardano Oct 10 '14 at 17:40
  • $\begingroup$ @G.Cardano You're welcome. $\endgroup$ – Random Variable Oct 10 '14 at 19:04
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Hints:

Substitute $x=\frac{t}{1-t}$, followed by $t=1-u$, and then expand the denominator via partial fractions:

$$\begin{align} \mathcal{I} &=\int_{0}^{1}\frac{\ln^3{\left(1+x\right)}}{x}\,\mathrm{d}x\\ &=\int_{0}^{\frac12}\frac{(1-t)\ln^3{\left(\frac{1}{1-t}\right)}}{t}\,\frac{1}{(1-t)^2}\mathrm{d}t\\ &=-\int_{0}^{\frac12}\frac{\ln^3{\left(1-t\right)}}{t(1-t)}\,\mathrm{d}t\\ &=-\int_{\frac12}^{1}\frac{\ln^3{\left(u\right)}}{u(1-u)}\,\mathrm{d}u\\ &=-\int_{\frac12}^{1}\frac{\ln^3{\left(u\right)}}{u}\,\mathrm{d}u-\int_{\frac12}^{1}\frac{\ln^3{\left(u\right)}}{1-u}\,\mathrm{d}u\\ &=-\left[\frac{\ln^4{(u)}}{4}\right]_{\frac12}^{1}-\int_{\frac12}^{1}\frac{\ln^3{\left(u\right)}}{1-u}\,\mathrm{d}u\\ &=\frac{\ln^4{(2)}}{4}-\int_{\frac12}^{1}\frac{\ln^3{\left(u\right)}}{1-u}\,\mathrm{d}u.\\ \end{align}$$

The final integral $\int_{\frac12}^{1}\frac{\ln^3{\left(u\right)}}{1-u}\,\mathrm{d}u$ is hardly trivial, but neither is it terribly difficult.

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  • $\begingroup$ David H, thank you very much! +1 $\endgroup$ – G. Cardano Oct 10 '14 at 17:38
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First note that \begin{align} \int \frac{\ln^{2}(1+x)}{x} \, dx = - 2 \, Li_{3}(1+x) + 2 \, Li_{2}(1+x) \, \ln(1+x) + \ln(-x) \, \ln^{2}(1+x) \end{align} for which \begin{align} I_{2} = \int_{0}^{1} \frac{\ln^{2}(1+x)}{x} \, dx = \frac{\zeta(3)}{4}. \end{align}

Now, \begin{align} \int \frac{\ln^{3}(1+x)}{x} \, dx = \int \ln(1+x) \, \frac{\ln^{2}(1+x)}{x} \, dx \end{align} can be integrated by parts. This leads to \begin{align} \int \frac{\ln^{3}(1+x)}{x} \, dx &= \ln(1+x) \left( - 2 \, Li_{3}(1+x) + 2 \, Li_{2}(1+x) \, \ln(1+x) + \ln(-x) \, \ln^{2}(1+x) \right) \\ & - \int \frac{- 2 \, Li_{3}(1+x) + 2 \, Li_{2}(1+x) \, \ln(1+x) + \ln(-x) \, \ln^{2}(1+x)}{1+x} \, dx \\ &= \ln(1+x) \left( - 2 \, Li_{3}(1+x) + 2 \, Li_{2}(1+x) \, \ln(1+x) + \ln(-x) \, \ln^{2}(1+x) \right) \\ & \hspace{10mm} + 6 Li_{4}(1+x) + Li_{2}(1+x) \, \ln^{2}(1+x) - 4 Li_{3}(1+x) \, \ln(1+x) \end{align} This leads to \begin{align} I_{3} &= \int_{0}^{1} \frac{\ln^{3}(1+x)}{x} \, dx \\ &= 6 Li_{4}(2) + \ln^{2}(2) \, Li_{2}(2) - 4 \ln(2) \, Li_{3}(2) - 6 \zeta(4) - \frac{7}{4} \, \ln(2) \, \zeta(3) \\ &= \frac{\pi^{4}}{15} + \frac{\pi^{2}}{4} \, \ln^{2}(2) - \frac{21}{4} \, \zeta(3) \, \ln(2) - \frac{1}{4} \, \ln^{4}(2) - 6 Li_{4}\left(\frac{1}{2} \right) \end{align}

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  • $\begingroup$ Leucippus, thank you very much! +1. $\endgroup$ – G. Cardano Oct 10 '14 at 17:39
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do you know this results ?

$\int_0^1 \frac{\left(\log (1+x)\right)^2}{x}dx=(ζ(3))/4$

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    $\begingroup$ you could also try to explain why, so the OP understands the solving process. $\endgroup$ – cjferes Oct 9 '14 at 18:27
  • $\begingroup$ @DavidH: the value is now correct, but without explanation, not very useful. $\endgroup$ – robjohn Oct 9 '14 at 19:30

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