What is int_0^pi (lnx)^2∫π0(lnx)2?

1 Answer
Cesareo R.
Jul 8, 2016

int_0^pi (log_e x)^2dx=2 pi - 2 pi Log_e pi + pi (log_e pi)^2∫π0(logex)2dx=2π−2πlogeπ+π(logeπ)2

Explanation:

d/(dx)(x cdot (log_e x)^2) = (log_e x)^2-2log_e xddx(x⋅(logex)2)=(logex)2−2logex

then

int (log_e x)^2dx = x cdot (log_e x)^2-2int log_e x dx = x((log_e x)^2-2(log_e x -1))∫(logex)2dx=x⋅(logex)2−2∫logexdx=x((logex)2−2(logex−1))

now

lim_{x->0}x((log_e x)^2-2(log_e x -1)) = 0

because

lim_{x->0}xlog_e x = x sum_{k=1}^{oo}(-1)^k/k (x-1)^k = 0

and

lim_{x->pi}x((log_e x)^2-2(log_e x -1)) = 2 pi - 2 pi Log_e pi + pi (log_e pi)^2

so finally

int_0^pi (log_e x)^2dx=2 pi - 2 pi Log_e pi + pi (log_e pi)^2

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