# Show that $\int_0^1 \frac{\ln^3(x)}{1+x} dx = \sum_{k=0}^{\infty}(-1)^{k+1} \frac6{(k+1)^4}$

Show that;

$$\int_0^1 \frac{\ln^3(x)}{1+x} dx = \sum_{k=0}^{\infty}(-1)^{k+1} \frac6{(k+1)^4}$$

I arrived to the fact that

$$\int_0^1 \frac{\ln^3(x)}{1+x} dx = \sum_{k=0}^{\infty}(-1)^k\int_0^1 x^k \ln^3(x)dx$$

But I am unable to continue further.

• Hint: integration by parts. – Marco Cantarini Aug 12 '16 at 18:28
• I don't know if you are interested but that series has also a nice closed form. – Marco Cantarini Aug 12 '16 at 18:33

An alternative approach is to use differentiation under the integral sign. For any $\alpha > 0$ we have $$\int_{0}^{1}\frac{x^{\alpha}}{1+x}\,dx = \sum_{n\geq 0}(-1)^n\int_{0}^{1}x^{\alpha+n}\,dx = \sum_{n\geq 0}\frac{(-1)^n}{n+\alpha+1}\tag{1}$$ hence by differentiating three times wrt $\alpha$ both sides of $(1)$ we get: $$\int_{0}^{1}\frac{x^{\alpha}\log^3(x)}{1+x}\,dx = \sum_{n\geq 0}\frac{6(-1)^{n+1}}{(n+\alpha+1)^4}\tag{2}$$ and by considering the limit s $\alpha\to 0^+$: $$\int_{0}^{1}\frac{\log^3(x)}{1+x}\,dx = 6\sum_{n\geq 1}\frac{(-1)^n}{n^4} = -\frac{21}{4}\zeta(4)=\color{red}{-\frac{7\pi^4}{120}}\tag{3}$$ follows.

Just integrate by parts:

\begin{align*} \int_0^1 x^k \ln^3(x) \, dx &= \frac{x^{k + 1}}{k + 1} \ln^3 x \big|_0^1 - \int_0^1 \frac{x^{k + 1}}{k + 1} 3 \ln^2(x) \frac 1 x \, dx \\ &= -\frac{3}{k + 1} \int_0^1 x^k \ln^2(x) \, dx \end{align*}

The next application reverses the sign, picks up factor $2/(k + 1)$, and drops the power on the log by $1$. Continue until the log disappears.

The substitution $u=-\ln x$ rewrites the integral as $$-\int_0^\infty \frac{u^3e^{-u}du}{1+e^{-u}}=\sum_{k=0}^\infty\left(-1\right)^{k+1}\int_0^\infty u^3e^{-\left(k+1\right)u}du.$$ The substitution $v=\left(k+1\right)u$ in $\int_0^\infty v^3 e^{-v}dv=3!=6$ obtains $$\int_0^\infty u^3e^{-\left(k+1\right)u}du=\frac{6}{\left(k+1\right)^4},\,\int_0^1\frac{\ln^3x}{1+x}dx=6\sum_{k=0}^\infty\frac{\left(-1\right)^{k+1}}{\left(k+1\right)^4}.$$

We begin with the substitution $$x=\mathrm{e}^{-y}$$ so our integral becomes $$-\int\limits_{0}^{\infty} \frac{y^{3}}{1+\mathrm{e}^{\,y}} \mathrm{d} y$$

This integral is a Mellin transform of the function $$f(y) = \frac{1}{1+\mathrm{e}^{\,y}}$$ where the Mellin transform is defined as $$\mathcal{M}[f(x](s) = \int\limits_{0}^{\infty} x^{s-1} f(x) \mathrm{d} x$$

From Tables of Integral Transforms (Bateman Manuscript) Volume 1, we have $$\mathcal{M}[(\mathrm{e}^{\alpha x} + 1)^{-1}](s) = \frac{1}{\alpha^{s}} \Gamma(s) (1-2^{1-s}) \zeta(s) \label{eq:1} \tag{1}$$

With $$s = 4 \quad \mathrm{and} \quad \alpha = 1$$ we have $$-\int\limits_{0}^{\infty} \frac{y^{3}}{1+\mathrm{e}^{y}} \mathrm{d} y = -\Gamma(4) (1-2^{1-4}) \zeta(4) = -\frac{7\pi^{4}}{120}$$

Thus $$\int\limits_{0}^{1} \frac{\mathrm{ln}^{3}(x)}{1+x} \mathrm{d} x = -\frac{7\pi^{4}}{120}$$

We can rewrite \eqref{eq:1} as \begin{align} \mathcal{M}[(\mathrm{e}^{\alpha x} + 1)^{-1}](s) & = \frac{1}{\alpha^{s}} \Gamma(s) (1-2^{1-s}) \zeta(s) \\ & = \frac{1}{\alpha^{s}} \Gamma(s) \eta(s) \end{align} where $$\eta(s) = (1-2^{1-s}) \zeta(s) = \sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{s}}$$ is the Dirichlet eta function, also known as the alternating Riemann zeta function.

Note that for Wolfram Alpha, $\eta(s)$ defaults to the Dedekind eta function. To obtain the Dirichlet eta function type "dirichlet eta(s)".

Substitute $x=e^{-u}$: \begin{align} \int_0^1\frac{\log^3(x)}{1+x}\,\mathrm{d}x &=-\int_0^\infty\frac{u^3}{1+e^{-u}}e^{-u}\,\mathrm{d}u\\ &=\int_0^\infty\sum_{k=1}^\infty(-1)^ku^3e^{-ku}\,\mathrm{d}u\\ &=\int_0^\infty u^3e^{-u}\sum_{k=1}^\infty\frac{(-1)^k}{k^4}\,\mathrm{d}u\\ &=3!\sum_{k=1}^\infty\frac{(-1)^k}{k^4}\\ \end{align}

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \color{#f00}{\int_{0}^{1}{\ln^{3}\pars{x} \over 1 + x}\,\dd x} & = \lim_{\mu \to 0}\,\partiald[3]{}{\mu}\int_{0}^{1}{x^{\mu} \over 1 + x}\,\dd x = \lim_{\mu \to 0}\,\partiald[3]{}{\mu} \int_{0}^{1}{x^{\mu} - x^{\mu + 1} \over 1 - x^{2}}\,\dd x \\[5mm] & \stackrel{x^{2}\ \mapsto\ x}{=}\ \half\,\lim_{\mu \to 0}\,\partiald[3]{}{\mu}\bracks{% \int_{0}^{1}{x^{\mu/2 - 1/2}\,\,\, -\,\,\, x^{\mu/2} \over 1 - x}\,\dd x} \\[5mm] & = \half\,\lim_{\mu \to 0}\,\partiald[3]{}{\mu}\bracks{% \int_{0}^{1}{1 - x^{\mu/2} \over 1 - x}\,\dd x - \int_{0}^{1}{1 - x^{\mu/2 - 1/2} \over 1 - x}\,\dd x} \\[5mm] & = \half\,\lim_{\mu \to 0}\,\partiald[3]{}{\mu} \bracks{\Psi\pars{\mu + 2 \over 2} - \Psi\pars{\mu + 1 \over 2}}\tag{1} \end{align} where $\ds{\Psi}$ is the Digamma Function. In $\ds{\pars{1}}$ we used the identity: $\ds{\Psi\pars{\xi} = -\gamma + \int_{0}^{1}{1 - t^{\xi - 1} \over 1 - t} \,\dd \xi}$ with $\ds{\Re\pars{\xi} > 0}$.

\begin{align} \color{#f00}{\int_{0}^{1}{\ln^{3}\pars{x} \over 1 + x}\,\dd x} & = {1 \over 16}\bracks{\Psi'''\pars{1} - \Psi'''\pars{\half}} = \color{#f00}{-\,{7 \over 120}\,\pi^{4}} \approx -5.6822 \end{align}

Some values of $\ds{\Psi'''}$ are well known: $\ds{\Psi'''\pars{1} = 3! \times \zeta\pars{4} = {1 \over 15}\,\pi^{4}\,,\qquad \Psi'''\pars{\half} = 3! \times 15 \times \zeta\pars{4} = \pi^{4}}$

with $\ds{\zeta\pars{4} = {1 \over 90}\,\pi^{4}}$.