How do you determine if the improper integral converges or diverges #int ln(sin(x))# from 0 to pi/2?

1 Answer
Feb 7, 2018

This integral converges to #-pi/2Ln2#

Explanation:

#I=int_0^(pi/2) Ln(sinx)*dx#

=#[xLn(sinx)]_0^(pi/2)-int_0^(pi/2) xcotx*dx#

=#0-int_0^(pi/2) (x*dx)/tanx#

=#-int_0^(pi/2) (arctan(tanx)*dx)/tanx#

After using Feynman's trick,

#I(b)=-int_0^(pi/2) (arctan(btanx)*dx)/tanx#

#I'(b)=-int_0^(pi/2) dx/[(btanx)^2+1]#

=#-int_0^(pi/2) ([(tanx)^2+1]*dx)/([(btanx)^2+1][(tanx)^2+1])#

After using #u=tanx# and #du=[(tanx)^2+1]*dx# transformation, this integral became,

#I'(b)=-int_0^(oo) (du)/[((bu)^2+1)*(u^2+1)]#

=#1/(b^2-1)int_0^(oo) (du)/(u^2+1)#-#b^2/(b^2-1)int_0^(oo) (du)/((bu)^2+1)#-

=#1/(b^2-1)[arctanu]_0^(oo)#-#b/(b^2-1)[arctan(u/b)]_0^(oo)#

=#pi/2*[1/(b^2-1)-b/(b^2-1)]#

=#-pi/2*(b-1)/((b-1)*(b+1))#

=#-pi/2*1/(b+1)#

After integrating both sides,

#I(b)=-pi/2Ln(b+1)+C#

For vanishing original integral, set #b=0#

#I(0)=-pi/2Ln(0+1)+C#

#C-pi+2*Ln1=0#

#C=pi/2*0=0#

Thus, #I=I(1)=-pi/2Ln2#