0-1-ln-x-2-1-x-2-dx-pi-3-16-

Question Number 97369 by M±th+et+s last updated on 07/Jun/20

\int_{0}^{1} \frac{{(l n (x))}^{2}}{1 + x^{2}} d x = \frac{π^{3}}{16}

Answered by maths mind last updated on 07/Jun/20

h e l l o h a p p y t o c o m b a c k

\int_{0}^{1} \frac{{l n}^{2} (z)}{1 + z^{2}} d z = \int_{1}^{\infty} \frac{{l n}^{2} (z)}{1 + z^{2}} d z

\Rightarrow \int_{0}^{1} \frac{{l n}^{2} (z)}{1 + z^{2}} d z = \frac{1}{2} \int_{0}^{\infty} \frac{{l n}^{2} (z)}{1 + z^{2}} d z

\int_{- \infty}^{+ \infty} \frac{{l n}^{2} (z)}{1 + z^{2}} d z = 2 i π R e s (\frac{{l n}^{2} (z)}{1 + z^{2}}, z = i)

\Rightarrow \int_{0}^{+ \infty} \frac{{l n}^{2} (z)}{1 + z^{2}} d z + \int_{- \infty}^{0} \frac{{l n}^{2} (z)}{1 + z^{2}} d z

= \int_{0}^{+ \infty} \frac{{l n}^{2} (z) + {l n}^{2} (z) - π^{2} + 2 i π l n (z)}{1 + z^{2}} d z = 2 i π . \frac{{l n}^{2} (i)}{2 i} = π . - \frac{π^{2}}{4}

\Leftrightarrow 2 \int_{0}^{+ \infty} \frac{{l n}^{2} (z)}{1 + z^{2}} d z - π^{2} \int_{0}^{+ \infty} \frac{d z}{1 + z^{2}} = - \frac{π^{3}}{4}

\Rightarrow 2 \int_{0}^{+ \infty} \frac{{l n}^{2} (z)}{1 + z^{2}} d z - \frac{π^{3}}{2} = - \frac{π^{3}}{4} \Leftrightarrow \int_{0}^{+ \infty} \frac{{l n}^{2} (z)}{1 + z^{2}} = 2 \int_{0}^{1} \frac{{l n}^{2} (z)}{1 + z^{2}} d z = \frac{π^{3}}{8}

\Rightarrow \int_{0}^{1} \frac{{l n}^{2} (z)}{1 + z^{2}} d z = \frac{π^{3}}{16}

Commented by M±th+et+s last updated on 07/Jun/20

h o w a r e y o u s i r w e l l l l l l c o m e e e e e b a c k

v e r y h a p p y t o s e e y o u a g a i n

Commented by maths mind last updated on 07/Jun/20

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