Integrate \(\int_{0}^{1} \frac{\mathrm{ln}^{3}(x)}{1+x} \mathrm{d} x\)

The following integral was a question on Mathematics Stack Exchange

\begin{equation}
\int\limits_{0}^{1} \frac{\mathrm{ln}^{3}(x)}{1+x} \mathrm{d} x
\label{eq:160812a1}
\tag{1}
\end{equation}

We begin with the substitution \(x=\mathrm{e}^{-y}\) so our integral becomes
\begin{equation}
-\int\limits_{0}^{\infty} \frac{y^{3}}{1+\mathrm{e}^{\,y}} \mathrm{d} y
\label{eq:160812a2}
\tag{2}
\end{equation}

This integral is a Mellin transform of the function
\begin{equation}
f(y) = \frac{1}{1+\mathrm{e}^{\,y}}
\label{eq:160812a3}
\tag{3}
\end{equation}
where the Mellin transform is defined as
\begin{equation}
\mathcal{M}[f(x)](s) = \int\limits_{0}^{\infty} x^{s-1} f(x) \mathrm{d} x
\label{eq:160812a4}
\tag{4}
\end{equation}

From Tables of Integral Transforms (Bateman Manuscript) Volume 1, we have
\begin{equation}
\mathcal{M}[(\mathrm{e}^{\alpha x} + 1)^{-1}](s) = \frac{1}{\alpha^{s}} \Gamma(s) (1-2^{1-s}) \zeta(s)
\label{eq:160812a5}
\tag{5}
\end{equation}

With \(s = 4\) and \(\alpha = 1\) we have
\begin{equation}
-\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}
\label{eq:160812a6}
\tag{6}
\end{equation}

Thus
\begin{equation}
\int\limits_{0}^{1} \frac{\mathrm{ln}^{3}(x)}{1+x} \mathrm{d} x = -\frac{7\pi^{4}}{120}
\label{eq:160812a7}
\tag{7}
\end{equation}

Addendum

We can rewrite \eqref{eq:160812a5} 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)
\label{eq:160812a8}
\tag{8}
\end{align}
where
\begin{equation}
\eta(s) = (1-2^{1-s}) \zeta(s) = \sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{s}}
\end{equation}
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)”.

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