What is the integral of #1/ln(lnx)#?

Question

Christopher Agresta · Accepted Answer

This is an impossible integral to complete without resorting to calculators and evaluating at explicit bounds. https://tutor.hix.ai Let's see how far we can go, though... #int (ln(lnx))^(-1)dx# Let:
#u = lnx#
#du = 1/xdx#
#xdu = dx -> e^udu = dx# #= int (lnu)^(-1)e^udu# #= inte^u/(lnu)du# #st - int t ds# Let:
#s = e^u#
#ds = e^udu#
#dt = 1/lnudu#
#t = ?# Detour. We need to know the integral for #1/lnx#. Let:
#q = 1/lnu#
#dq = -lnu/udu#
#dr = du#
#r = u# #qr - int r dq# #= u/lnu + int lnudu# #= u/lnu + u ln|u| - u = t# Back to #s# and #t#: #= e^u (u/lnu + u ln|u| - u) - int e^u(u/lnu + u lnu - u)du# #= (ue^u)/lnu + ue^u ln|u| - u e^u - int (ue^u)/lnudu - int ue^u lnudu + intu e^udu# The only integral we can do with real functions or standard functions is #int u e^udu#. I'm running out of variables. #op - int p do# Let:
#o = u#
#do = du#
#dp = e^udu#
#p = e^u# #ue^u - int e^udu# #= ue^u - e^u# So now we get: #= (ue^u)/lnu + ue^u ln|u| cancel(- u e^u + ue^u) - e^u - int (ue^u)/lnudu - int ue^u lnudu# #= (ue^u)/lnu + ue^u ln|u| - e^u - int (ue^u)/lnudu - int ue^u lnudu# #= e^u[u/lnu + u ln|u| - 1] - int (ue^u)/lnudu - int ue^u lnudu# #= e^(lnx)[(lnx)/ln(lnx) + (lnx) ln|lnx| - 1] - int ((lnx)e^(lnx))/(xln(lnx))dx - int 1/x(lnx)e^(lnx) ln(lnx)dx# #= color(blue)(x[(lnx)/ln(lnx) + (lnx) ln|lnx| - 1] - int (lnx)/(ln(lnx))dx - int (lnx) ln(lnx)dx)#

Charles Anctil · Answer

undefined The integral of $ \frac{1}{\ln(\ln(x))} $ with respect to $ x $ does not have a elementary closed-form expression in terms of standard mathematical functions. It is denoted by $ \int \frac{1}{\ln(\ln(x))} \, dx $. However, it can be expressed in terms of a special function called the logarithmic integral function denoted by $ \text{li}(x) $. Therefore, the integral can be written as $ \text{li}(\ln(x)) + C $, where $ C $ is the constant of integration.