What is #int_0^pi (lnx)^2 / x^(1/2)#?

1 Answer
Dec 16, 2015

Answer:

#int_0^piln^2(x)/sqrt(x)dx = 2sqrt(pi)(ln^2(pi)-4ln(pi)+8)#

Explanation:

First let's solve the indefinite integral #intln^2(x)/sqrt(x)dx#. by applying integration by parts twice:

First Integration by Parts

Let #u = ln^2(x)# and #dv = 1/sqrt(x)dx#

Then #du = (2ln(x))/x# and #v = 2sqrt(x)#

Thus

#intln^2(x)/sqrt(x)dx = intudv#

#= uv - intvdu#

#=2sqrt(x)ln^2(x) - 4intln(x)/sqrt(x)dx#

Second Integration by Parts

Let #u = ln(x)# and #dv = 1/sqrt(x)#

Then #du = 1/x# and #v = 2sqrt(x)#

Thus

#intln(x)/sqrt(x)dx = intudv#

#= uv - intvdu#

#= 2sqrt(x)ln(x) - 2int1/sqrt(x)dx#

#= 2sqrt(x)ln(x) - 4sqrt(x) + C#

Putting it together, we get

#intln^2(x)/sqrt(x)dx = 2sqrt(x)ln^2(x) - 4intln(x)/sqrt(x)dx#

#= 2sqrt(x)ln^2(x) - 4(2sqrt(x)ln(x) - 4sqrt(x) + C)#

#= 2sqrt(x)(ln^2(x) - 4ln(x) + 8) + C#

Now we can evaluate the original definite integral.

#int_0^piln^2(x)/sqrt(x)dx = [2sqrt(x)(ln^2(x) - 4ln(x) + 8)]_0^pi#

(Note that as there is a discontinuity at #0# we must evaluate this using a limit)

#= 2sqrt(pi)(ln^2(pi) - 4ln(pi) + 8) - lim_(x->0)2sqrt(x)(ln^2(x) - 4ln(x) + 8)#

#= 2sqrt(pi)(ln^2(pi) - 4ln(pi) + 8) - 0#

#= 2sqrt(pi)(ln^2(pi) - 4ln(pi) + 8)#