$\int_0^1 \left(\ln(x^{-1})\right)^2 dx$ =? How to evaluate this integral?
$$\int_0^1 \left(\ln(x^{-1})\right)^2 dx$$
I tried $u$-substitution but that did not work. I am not even sure if this integral converges.
 A: Since for any $s\in\left[-\frac{1}{2},\frac{1}{2}\right]$ we have
$$ \int_{0}^{1}x^s\,dx = \frac{1}{s+1}\tag{1} $$
by differentiation under the integral sign we have
$$ \frac{d^2}{ds^2}\int_{0}^{1}x^s\,dx = \color{green}{\int_{0}^{1}x^s\log^2(x)\,dx} = \frac{d^2}{ds^2}\left(\frac{1}{s+1}\right)=\color{green}{\frac{2}{(s+1)^3}}\tag{2}$$
and by evaluating $(2)$ at $s=0$
$$ \int_{0}^{1}\log^2(x)\,dx = \color{green}{\large 2}\tag{3} $$
follows.
A: Apply Integration By Parts:
$$u=\ln ^2\left(\frac{1}{x}\right),\:u'=\frac{2\ln \left(x\right)}{x},\:\:v'=1,\:\:v=x$$
$$\int _0^1\ln ^2\left(x^{-1}\right)dx=x\ln ^2\left(\frac{1}{x}\right)-\int \:2\ln \left(x\right)dx$$
$$\int \:2\ln \left(x\right)dx = \color{red}{2\left(x\ln \left(x\right)-x\right)}$$
So
$$\int _0^1\ln ^2\left(x^{-1}\right)dx=x\ln ^2\left(\frac{1}{x}\right)-2\left(x\ln \left(x\right)-x\right)=\color{red}{2}$$
A: Using integration by parts, let $a=\mathrm{ln}(x)$ and $\mathrm{d}b =\mathrm{ln}(x) \mathrm{d}x$
\begin{align}
\int\limits_{0}^{1} \mathrm{ln}^{2}(x) \mathrm{d}x &=
x \mathrm{ln}^{2}(x) - x\mathrm{ln}(x) \Big|_{0}^{1} - \int\limits_{0}^{1} (\mathrm{ln}(x) - 1) \mathrm{d}x \\
&= x \mathrm{ln}^{2}(x) - x\mathrm{ln}(x) - x\mathrm{ln}(x) +2x \Big|_{0}^{1} \\
&= 2
\end{align}
A: Let $x = e^{-t}$, $dx = - e^{-t} \, dt$ to obtain
\begin{align}
\int_{0}^{1} \ln^{2}\left(\frac{1}{x}\right) \, dx &= \int_{0}^{\infty} e^{-t} \, \ln^{2}(e^{t}) \, dt \\
&= \int_{0}^{\infty} t^{2} \, e^{-t} \, dt \\
&= [- (t^{2} + 2 t + 2) \, e^{-t} ]_{0}^{\infty} \\
&= 2.
\end{align}
