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

1 Answer
Jul 8, 2016

Answer:

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

Explanation:

#d/(dx)(x cdot (log_e x)^2) = (log_e x)^2-2log_e x#

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))#

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#