How would you integrate #int_1^e 1/(x sqrt(ln^2x))dx# ?

#int_1^e 1/(x sqrt(ln^2x))dx#

2 Answers
Mar 1, 2018

This integral does not exist.

Explanation:

Since #ln x>0# in the interval #[1,e]#, we have

#sqrt{ln^2 x}= |ln x| = ln x#

here, so that the integral becomes

#int_1^e dx/{x ln x}#

Substitute #ln x = u#, then #dx/x = du# so that

#int_1^e dx/{x ln x} = int_{ln 1}^{ln e} {du}/u = int_0^1 {du}/u#

This is an improper integral, since the integrand diverges at the lower limit. This is defined as

#lim_{l -> 0^+} int_l^1 {du}/u #

if this exists. Now

#int_l ^1 {du}/u = ln 1 - ln l = -ln l#

since this diverges in the limit #l -> 0^+#, the integral does not exist.

Mar 1, 2018

#pi/2#

Explanation:

The integral #int_1^e("d"x)/(xsqrt(1-ln^2(x))#.

Substitute first #u=ln(x)# and #"d"u=("d"x)/x#.

Thus, we have
#int_(x=1)^(x=e)("d"u)/sqrt(1-u^2)#

Now, substitute #u=sin(v)# and #"d"u=cos(v)"d"v#.

Then,
#int_(x=1)^(x=e)(cos(v))/(sqrt(1-sin^2(v)))\ "d"v=int_(x=1)^(x=e)"d"v# since #1-sin^2(v)=cos^2(v)#.

Continuing, we have
#[v]_(x=1)^(x=e)=[arcsin(u)]_(x=1)^(x=e)=[arcsin(ln(x))]_(x=1)^(x=e)=arcsin(ln(e))-arcsin(ln(1))=pi/2-0=pi/2#