Calculus Question: $\int_2^\infty\frac{\log^3(x-1)}{x^2}dx$ I have just taken calculus quiz but I could not find  $\displaystyle \int_2^\infty\frac{\log^3(x-1)}{x^2}dx$? Any help would be appreciated. Thanks in advance.
EDIT:
Forgot to mention, my tutor gave us hints about this question.


*

*Use Taylor series

*$\displaystyle \zeta(3)=\sum_{n=1}^{\infty}\frac{1}{n^3}$


Those the hints that she gave to us.
 A: Use the substitution $x-1=t$ to obtain:
$$\int_1^{\infty} \frac{\log^3t}{(1+t)^2}\,dt$$
With the substitution $t=1/u$, the above integral is:
$$\int_0^1 \frac{-\log^3 u}{(1+u)^2}\,du$$
Next, use the following series representation,
$$\frac{1}{(1+u)^2}=\sum_{n=0}^{\infty}(-1)^n (n+1)u^n$$
to obtain:
$$\int_0^1 \frac{-\log^3 u}{(1+u)^2}\,du=-\sum_{n=0}^{\infty} (-1)^n(n+1)\int_0^1 \log^3u \,u^n\,du$$
Next use the substitution $\log u=-x$ to get:
$$\sum_{n=0}^{\infty} (-1)^n(n+1)\int_0^{\infty} x^3e^{-(n+1)x}\,dx$$
With another substitution $(n+1)x=y$, you should get:
$$\sum_{n=0}^{\infty} \frac{(-1)^n}{(n+1)^3}\int_0^{\infty} y^3e^{-y}\,dy$$
Recoginse that $\int_0^{\infty} y^3e^{-y}\,dy=\Gamma(4)=3!$ (you can show this using integration by parts if you don't like the Gamma function), hence
$$\sum_{n=0}^{\infty} \frac{(-1)^n}{(n+1)^3}\int_0^{\infty} y^3e^{-y}\,dy=6\sum_{n=0}^{\infty} \frac{(-1)^n}{(n+1)^3}$$
It is easy to show that:
$$\sum_{n=0}^{\infty} \frac{(-1)^n}{(n+1)^3}=\frac{3\zeta(3)}{4}$$
Hence, our final answer is:
$$\boxed{\dfrac{9\zeta(3)}{2}}$$
A: By substitutions, the following integrals are equivalent:
\begin{align*}
  \int_{2}^{\infty} \, \frac{\log^3(x-1)}{x^2}\, dx &= \int_{1}^{\infty} \, \frac{\log^3(x)}{(1+x)^2} \, dx\\
&= -\int_{0}^{1} \, \frac{\log^3(x)}{(1+x)^2}\, dx \tag 1
\end{align*}
$(1)$ can be written as a sum, consider:
\begin{align*}
  \int_{0}^{1} \, \frac{x^a}{(1+x)^2} dx &= \int_{0}^{1} \, \sum_{n\ge 0} (-1)^n (n+1)\, x^{a+n} \, dx\\
&= \sum_{n\ge 0} \int_{0}^{1} \, (-1)^n (n+1)\, x^{a+n} \, dx\\
&= \sum_{n\ge 0} (-1)^n \frac{(n+1)}{a+n+1}\tag 2
\end{align*}
Differentiate $(2)$ w.r.t a thrice and set $a=0$, and from $(1)$,
\begin{align*}
  \int_{2}^{\infty} \, \frac{\log^3(x-1)}{x^2}\, dx &= 6\, \sum_{n\ge 0}\frac{(-1)^n}{(n+1)^3}\\
&= \frac{9}{2} \zeta{(3)} \approx 5.40925606421817428
\end{align*}
So, a general result looks like:
\begin{align*}
  \int_{2}^{\infty} \, \frac{\log^n(x-1)}{x^2}\, dx &= \left(1-\frac{1}{2^{n-1}}\right)n!\, \zeta{(n)}
\end{align*}
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{\int_{2}^{\infty}{\ln^{3}\pars{x - 1} \over x^{2}}\,\dd x}$

\begin{align}&\color{#c00000}{\int_{2}^{\infty}%
{\ln^{3}\pars{x - 1} \over x^{2}}\,\dd x}
=\int_{1}^{\infty}{\ln^{3}\pars{x} \over \pars{x + 1}^{2}}\,\dd x
=\int_{1}^{0}{\ln^{3}\pars{1/x} \over \pars{1/x + 1}^{2}}
\,\pars{-\,{\dd x \over x^{2}}}
\\[3mm]&=\color{#00f}{-\int_{0}^{1}{\ln^{3}\pars{x} \over \pars{1 + x}^{2}}\,\dd x}
\end{align}

With $\ds{0 < \epsilon < 1}$:
\begin{align}&\color{#00f}{-\int_{\epsilon}^{1}%
{\ln^{3}\pars{x} \over \pars{1 + x}^{2}}\,\dd x}
=-\,{\ln^{3}\pars{\epsilon} \over 1 + \epsilon}
-\int_{\epsilon}^{1}{1 \over 1 + x}\,\bracks{3\ln^{2}\pars{x}\,{1 \over x}}\,\dd x
\\[3mm]&=-\,{\ln^{3}\pars{\epsilon} \over 1 + \epsilon}
-3\int_{\epsilon}^{1}{\ln^{2}\pars{x} \over x}\,\dd x
+3\int_{\epsilon}^{1}{\ln^{2}\pars{x} \over 1 + x}\,\dd x
\\[3mm]&=-\,{\ln^{3}\pars{\epsilon} \over 1 + \epsilon} + \ln^{3}\pars{\epsilon}
+3\int_{\epsilon}^{1}{\ln^{2}\pars{x} \over 1 + x}\,\dd x
\end{align}

With the limit $\ds{\epsilon \to 0^{+}}$:
  \begin{align}&\color{#66f}{\large\int_{2}^{\infty}%
{\ln^{3}\pars{x - 1} \over x^{2}}\,\dd x}
=3\int_{0}^{1}{\ln^{2}\pars{x} \over 1 + x}\,\dd x
=-3\int_{0}^{1}\ln\pars{1 + x}\bracks{2\ln\pars{x}\,{1 \over x}}\,\dd x
\\[3mm]&=-6\int_{0}^{-1}\ln\pars{-x}\,{\ln\pars{1 - x} \over x}\,\dd x
=6\int_{0}^{-1}\ln\pars{-x}{\rm Li}_{2}'\pars{x}\,\dd x
=-6\int_{0}^{-1}{{\rm Li}_{2}\pars{x} \over x}\,\dd x
\\[3mm]&=-6\int_{0}^{-1}{\rm Li}_{3}'\pars{x}\,\dd x
=-6\,{\rm Li}_{3}\pars{-1}
=-6\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n^{3}}
\\[3mm]&=-6\bracks{\sum_{n = 1}^{\infty}{1 \over \pars{2n}^{3}}
-\sum_{n = 1}^{\infty}{1 \over \pars{2n - 1}^{3}}}
=-6\bracks{\sum_{n = 1}^{\infty}{1 \over \pars{2n}^{3}}
-\sum_{n = 1}^{\infty}{1 \over n^{3}}
+\sum_{n = 1}^{\infty}{1 \over \pars{2n}^{3}}}
\\[3mm]&=-6\braces{2\bracks{{1 \over 8}\sum_{n = 1}^{\infty}{1 \over n^{3}}}
-\sum_{n = 1}^{\infty}{1 \over n^{3}}}
={9 \over 2}\sum_{n = 1}^{\infty}{1 \over n^{3}}
=\color{#66f}{\large{9 \over 2}\,\zeta\pars{3}} \approx {\tt 5.4093}
\end{align}

$\ds{{\rm Li_{s}}\pars{z}}$ is the
PolyLogarithm Function and we used well known properties of them as explained in the above cited link.
A: There is another way to solve. From @Pranav Arora, we know
$$ \int_0^\infty\frac{\ln^3x}{(1+x)^2}dx=-\int_0^1\frac{\ln^3u}{(1+u)^2}du. $$
Let
$$ I(\alpha)=\int_0^1\frac{u^\alpha}{(1+u)^2}du. $$
Clearly
$$ I'''(0)=\int_0^1\frac{\ln^3u}{(1+u)^2}du. $$
Since
\begin{eqnarray*}
I(\alpha)&=&\int_0^1\frac{u^\alpha}{(1+u)^2}du=\int_0^1u^\alpha\sum_{n=0}^\infty(-1)^n(n+1)u^{n}du\\
&=&\int_0^1\sum_{n=0}^\infty(-1)^n(n+1)u^{n+\alpha}du\\
&=&\sum_{n=0}^\infty(-1)^n\frac{n+1}{n+\alpha+1}
\end{eqnarray*}
the rest is the same as @gar's answer.
