# What is int_0^pi (lnx)^2?

Jul 8, 2016

${\int}_{0}^{\pi} {\left({\log}_{e} x\right)}^{2} \mathrm{dx} = 2 \pi - 2 \pi L o {g}_{e} \pi + \pi {\left({\log}_{e} \pi\right)}^{2}$

#### Explanation:

$\frac{d}{\mathrm{dx}} \left(x \cdot {\left({\log}_{e} x\right)}^{2}\right) = {\left({\log}_{e} x\right)}^{2} - 2 {\log}_{e} x$

then

$\int {\left({\log}_{e} x\right)}^{2} \mathrm{dx} = x \cdot {\left({\log}_{e} x\right)}^{2} - 2 \int {\log}_{e} x \mathrm{dx} = x \left({\left({\log}_{e} x\right)}^{2} - 2 \left({\log}_{e} x - 1\right)\right)$

now

${\lim}_{x \to 0} x \left({\left({\log}_{e} x\right)}^{2} - 2 \left({\log}_{e} x - 1\right)\right) = 0$

because

${\lim}_{x \to 0} x {\log}_{e} x = x {\sum}_{k = 1}^{\infty} {\left(- 1\right)}^{k} / k {\left(x - 1\right)}^{k} = 0$

and

${\lim}_{x \to \pi} x \left({\left({\log}_{e} x\right)}^{2} - 2 \left({\log}_{e} x - 1\right)\right) = 2 \pi - 2 \pi L o {g}_{e} \pi + \pi {\left({\log}_{e} \pi\right)}^{2}$

so finally

${\int}_{0}^{\pi} {\left({\log}_{e} x\right)}^{2} \mathrm{dx} = 2 \pi - 2 \pi L o {g}_{e} \pi + \pi {\left({\log}_{e} \pi\right)}^{2}$