Exhausting integral $ \int_{0}^{\infty}\frac{\ln^3(x)}{(x-1)^3} \, \mathrm{d}x $ I would like to evaluate following integral:
$$ \int\limits_{0}^{\infty}\frac{\ln^3(x)}{(x-1)^3} \, \mathrm{d}x $$
by complex means. I integrate
$$f(z)=\frac{\ln^4(z)}{(z-1)^3}$$
(where ln is analytic in $\mathbb{C} \setminus \mathbb{R_+}$ with $\ln(-1)=i\pi$) over following contour: $[0;R]$, $C_0(R)$ - circle of radius $R$, $[R, 1+\varepsilon]$, semicircle in lower half-plane $C_1(\varepsilon)$ and $[1-\varepsilon,0]$. After evaluation and taking $\lim\limits_{\varepsilon \to 0}$ I came to correct answer, but my question is:

Is it possible to avoid that exhausting evaluation?

Thanks in advance.
 A: The computation can be made simpler if we use two versions of the complex logarithms:

*

*$\log_{-}(z)$ denotes the complex logarithm with $\arg \in (-2\pi, 0)$.

*$\log_{+}(z)$ denotes the complex logarithm with $\arg \in (0, 2\pi)$.

In particular, for each $x > 0$, we have
\begin{align*}
\log_-(x+i0^+) &= \log x - 2\pi i, &
\log_-(x-i0^+) &= \log x, \\
\log_+(x+i0^+) &= \log x, &
\log_+(x-i0^+) &= \log x + 2\pi i.
\end{align*}
Now define
$$ f(z) = \frac{\log^2_+(z) \log^2_-(z)}{(z-1)^3}. $$
This choice simplifies the computation in two ways:

*

*The symmetry allows us to better relate the contour integral of $f$ to the real integral in the question.


*Regardless of whether $z$ is above or below the $x$-axis, either $\log_+(z)^2$ or $\log_-(z)^2$ cancels out $(z-1)^2$ in the denominator. This is particularly helpful when computing the limit of the integral along the semicircle around $z=1$.
Now consider the modified keyhole contour $\mathcal{C}(\varepsilon,R)$ consisting of:


*

*$\mathsf{L}_{+,1} = [i0^+, 1-\varepsilon + i0^+]$,

*clockwise upper semicircle $\mathsf{C}_{+}(1,\varepsilon)$,

*$\mathsf{L}_{+,2} = [1+\varepsilon + i0^+, R + i0^+]$,

*counter-clockwise circle $\mathsf{C}(0,R)$,

*$\mathsf{L}_{-,2} = [R - i0^-, 1+\varepsilon - i0^+]$,

*clockwise lower semicircle $\mathsf{C}_{-}(1,\varepsilon)$,

*$\mathsf{L}_{-,1} = [1-\varepsilon + i0^-, i0^-]$.

Then it is easy to check that
\begin{align*}
&\int_{\mathsf{L}_{+,1} \cup \mathsf{L}_{-,1}} f(z) \, \mathrm{d}z + \int_{\mathsf{L}_{+,2} \cup \mathsf{L}_{-,2}} f(z) \, \mathrm{d}z \\
&= \int_{0}^{1-\varepsilon} \frac{(-8\pi i)\log^3 x}{(x-1)^3} \, \mathrm{d}x
+ \int_{1+\varepsilon}^{R} \frac{(-8\pi i)\log^3 x}{(x-1)^3} \, \mathrm{d}x.
\end{align*}
Moreover, since $(z-1)f(z)$ is bounded near $z = 1$,
$$ \lim_{\varepsilon \to 0^+} \int_{\mathsf{C}_{+}(1,\varepsilon)} f(z) \, \mathrm{d}z
= (-\pi i) \lim_{\substack{z \to 1 \\ \operatorname{Im}(z) > 0}} (z-1)f(z)
= (-\pi i) (-2\pi i)^2
= 4\pi^3 i $$
and similarly
$$ \lim_{\varepsilon \to 0^+} \int_{\mathsf{C}_{-}(1,\varepsilon)} f(z) \, \mathrm{d}z
= (-\pi i) \lim_{\substack{z \to 1 \\ \operatorname{Im}(z) < 0}} (z-1)f(z)
= (-\pi i) (2\pi i)^2
= 4\pi^3 i. $$
Finally, it is clear that
$$ \lim_{R \to \infty} \int_{\mathsf{C}(0, R)} f(z) \, \mathrm{d}z = 0. $$
Now, since $f$ is analytic on and inside of $\mathcal{C}(\varepsilon,R)$, it follows that
$$ 0 = \lim_{\substack{R \to \infty \\ \varepsilon \to 0^+}} \int_{\mathcal{C}(\varepsilon,R)} f(z) \, \mathrm{d}z
= (-8\pi i)\int_{0}^{\infty} \frac{\log^3 x}{(x-1)^3} \, \mathrm{d}x + 8\pi^3 i $$
and therefore
$$ \int_{0}^{\infty} \frac{\log^3 x}{(x-1)^3} \, \mathrm{d}x = \pi^2. $$
A: Here's a way to do it only using the fact that $\int_{0}^{\infty}\frac{x^{s-1}}{e^{ax}-1}dx=\frac{\Gamma(s)\zeta(s)}{a^s}$.
First break the integral in two intervals and perform a substitution $x\to 1/x$ and then $x\to e^x$
$$\int_{0}^{\infty}\frac{\ln^3(x)}{(x-1)^3}dx=\left(\int_{0}^{1}+\int_{1}^{\infty}\right)\frac{\ln^3(x)}{(x-1)^3}dx=\int_{1}^{\infty}\frac{(1+x)\ln^3(x)}{(x-1)^3}=2\int_0^{\infty}dt\frac{t^3 e^t}{(e^t-1)^3}+\int_0^{\infty}dt\frac{t^3 e^t}{(e^t-1)^2}\equiv I_1+I_2$$
We can compute the integrals $I_1,I_2$ as follows. We see that
$$\frac{\partial }{\partial a}\int_{0}^{\infty}\frac{t}{e^{at}-1}dt=-\int_{0}^{\infty}\frac{t^2}{e^{at}-1}dt-\int_{0}^{\infty}\frac{t^2}{(e^{at}-1)^2}dt$$
Taking another derivative
$$\frac{\partial^2 }{\partial a^2}\int_{0}^{\infty}\frac{t}{e^{at}-1}dt+\frac{\partial}{\partial a}\int_{0}^{\infty}\frac{t^2}{e^{at}-1}dt=2\int_{0}^{\infty}dt\frac{t^3 e^{at}}{(e^{at}-1)^3}=I(a)$$
Note that $I(a=1)=I_1$. We find that it is equal to
$$I(a)=\frac{\partial^2}{\partial a^2}\frac{\zeta(2)}{a^2}+2\frac{\partial }{\partial a}\frac{\zeta(3)}{a^3}=6\frac{\zeta(2)-\zeta(3)}{a^4}$$
Similarly we note that
$$\frac{\partial }{\partial a}\int_{0}^{\infty}\frac{t^2}{e^{at}-1}dt=-\int_{0}^{\infty}\frac{t^3e^{at}}{(e^{at}-1)^2}dt=-J(a)$$
where $J(1)=I_2$. We conclude that
$$J(a)=\frac{6\zeta(3)}{a^4}$$.
Adding it all together we obtain that
$$\int_{0}^{\infty}\frac{\ln^3(x)}{(x-1)^3}=6\zeta(2)=\pi^2$$
PS instead of differentiation under the integral sign one can also use smart integration by parts. I don't know if this is a simpler way to do it but it definitely requires way less complex analysis and only a few basic identities.
A: Integrate by parts
\begin{align}
\int_{0}^{\infty}\frac{\ln^3x}{(x-1)^3} {d}x =&
 - \frac12 \int_{0}^{\infty}\frac{\ln^3x}x d\left(\frac{x}{x-1} \right)^2\\
 =& \>\frac32 \int_{0}^{\infty}\frac{\ln^2x}{(x-1)^2} {d}x 
 - \frac12 \int_{0}^{\infty}\underset{=0}{\overset{x\to1/x}{\frac{\ln^3x}{(x-1)^2}}} {d}x \\
= & \>3\int_{0}^{1}\frac{\ln^2x}{(x-1)^2} {d}x  
=3 \int_0^1 \ln^2x\>d\underset{IBP} {\left( \frac x{1-x}\right)}\\
=& \>6\int_{0}^{1}\frac{\ln x}{x-1} {d}x  
= 6\left(\frac{\pi^2}6\right)=\pi^2
\end{align}
