How do you evaluate the integral #int 1/x dx# from 1 to #oo#?

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Answer 1 · 2016-08-21T03:06:41

The integral does not converge.

Note that #1/x# is the derivative of #ln(x)#, so the integral #int1/xdx=ln(x)+C#.

In this case:

#int_1^oo1/xdx=[ln(x)]_1^oo#

Now evaluating, and using a limit for infinity:

#=lim_(xrarroo)ln(x)-ln(1)#

#ln(1)=0# and the limit approaches infinity, thus the integral does not converge.

#=oo#

If you don't understand why #lim_(xrarroo)ln(x)=oo#, take a look at the graph of #ln(x)#:

graph{lnx [-11.39, 39.92, -12.47, 13.19]}

The function steadily rises.