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I would like to show the following identity:

$$\boxed{ I := \int_0^1 \dfrac{(\ln(x))^5}{1+x} \mathrm{d}x = -\dfrac{31\pi^6}{252} }$$

Here is what I tried. The change of variables $u=1/x$ yields $$I= \int_1^{\infty} \dfrac{(\ln(x))^5}{1+1/u} \dfrac{1}{u^2} \mathrm{d}u = \int_1^{\infty} \dfrac{(\ln(x))^5}{u^2+u} \mathrm{d}u$$ Then $z=u-1$ gives $$I = \int_{0}^{\infty} \dfrac{(\ln(z+1))^5}{z^2+3z+2} \mathrm{d}z $$ with $z^2+3z+2=(z+1)(z+2)$. I wanted to use contour integration like here, but I was not sure how to proceed in this case. Anyway, the computations of the residues (of which "well-chosen" function? Maybe something like this?) seem to be difficult.

I believe that we can generalize to $\frac{(\ln(x))^n}{1+x}$, or maybe even more (e.g. $\frac{(\ln(x))^n}{ax^2+bx+c}$). Related computations are: (1), (2), (3), (4).

Thank you for your detailed help.

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    Related: https://math.stackexchange.com/questions/2117032 – 2017-01-27

1 Answers 1

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HINT:

Enforce the substitution $x\to e^{-x}$, expand the resulting denominator in a geometric series of $\sum_{n=0}^{\infty}(-1)^ne^{-nx}$, interchange the sum and integral, carryout the integral by either successive IBP or differentiating under the integral, and evaluate the resulting series representation of $\zeta(6)$.

Alternatively, note that

$$\int_0^1 \frac{\log^5(x)}{1+x}\,dx=\left. \left(\frac{d^5}{da^5}\int_0^1\frac{x^a}{1+x}\,dx\right)\right|_{a=0}$$

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    I was just thinking I recall someone having a nice answer for this problem maybe a year or two ago here, and from your hint I'm willing to bet it was you. – 2017-01-27
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    OK, I think I've managed to solve it: $$ \int_0^{\infty} \dfrac{(-u)^5}{1+e^{-u}} e^{-u} du =\\ -\int_0^{\infty} u^5\left[1 - \dfrac{1}{1+e^{-u}} \right] du = \\ -\int_0^{\infty} u^5\left[ - \sum_{n \geq 1} (-1)^n e^{-nu} \right] du = \\ \sum_{n \geq 1} (-1)^n \int_0^{\infty} u^5 e^{-nu} du = \\ \sum_{n \geq 1} (-1)^n \dfrac{120}{n^6} = \\ -\dfrac{31\pi^6}{252} $$ – 2017-01-27
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    That is correct. Post the answer and I'll up vote. – 2017-01-27
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    The last equality comes from $\sum_{n \geq 1} (-1)^n n^{-k} = \zeta(k)\left(2^{1-k} - 1 \right)$ (like [here](http://math.stackexchange.com/questions/1169861)), and $\zeta(6) = \pi^6/945$, which is not an obvious result. But still this is a clever way! Thank you. – 2017-01-27