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.
Those the hints that she gave to us.
On
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*}
On
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.
On
$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{2}^{\infty}{\ln^{3}\pars{x - 1} \over x^{2}}\,\dd x}$
\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.
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: $$\boxed{\dfrac{9\zeta(3)}{2}}$$