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What is #int_-oo^oo 3/x dx#?

Answer 1

One might think of splitting this into two integrals at first:

#int_(-oo)^(0) 3/xdx + int_(0)^(oo) 3/xdx#

This integral is fairly easy as an indefinite integral if you recall the antiderivative of #1/x#; #int 1/xdx = ln|x|#. The challenge is evaluating the boundaries, if it is possible.

#=> |[3ln|x|]|_(-oo)^(0) + |[3ln|x|]|_(0)^(oo)#

#= ([3ln|0|] - lim_(x->-oo) ln|x|) + (lim_(x->oo) ln|x| - [3ln|0|])#

#= cancel(3ln|0|) - lim_(x->-oo) ln|x| + lim_(x->oo) ln|x| - cancel(3ln|0|)#

#= -3lim_(x->-oo) ln|x| + 3lim_(x->oo)ln|x|#

but since it is #|x|#, #lim_(x->oo) = lim_(x->-oo)#:

#= 0*lim_(x->oo)ln|x|#

#= 0*oo#

#=># does not converge

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Answer 2

Christopher Allers

The integral of ( \frac{3}{x} ) from ( -\infty ) to ( \infty ) diverges.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.