I was given a problem to evaluate the following integral, ∫ − 4 − 1 1 x d xI know the fact that integration of 1 / x is log ⁡ ( x ) but how can we compute negative logarithm? I mean to say that since definite integral refers to the area under the curve so how can it be even defined?I tried the following method: ∫ − 4 − 1 1 x d x = log ⁡ ( − 1 ) − log ⁡ ( − 4 ) ⟹ log ⁡ ( − 1 − 4 ) = − log ⁡ ( 4 )Is it correct? I don't think that the answer should be − log ⁡ ( 4 ) because there is no graph for the negative logarithm. Is there any need of complex logarithm?

Question

I was given a problem to evaluate the following integral,      ∫          −      4              −      1                  1      x        d  xI know the fact that integration of   1  /  x is   log  ⁡  (  x  ) but how can we compute negative logarithm? I mean to say that since definite integral refers to the area under the curve so how can it be even defined?I tried the following method:      ∫          −      4              −      1                  1      x        d  x  =  log  ⁡  (  −  1  )  −  log  ⁡  (  −  4  )    ⟹    log  ⁡      (                            −          1                          −          4                      )    =  −  log  ⁡  (  4  )Is it correct? I don&#039;t think that the answer should be   −  log  ⁡  (  4  ) because there is no graph for the negative logarithm. Is there any need of complex logarithm?

Nia Molina · Accepted Answer

Negative logarithms do exist, just not in the real numbers. The complex log function yields the same value as the log function, but on negative numbers it adds a complex component, but which is constant!As a commenter noted, most books put the integral of       1    x   as   ln  ⁡  (  |  x  |  ) to prevent going outside the domain. However, if you allow for complex functions,   ln  ⁡   (  x  ) actually still works. In fact, they are equivalent, if you allow C to be complex, and allow for the fact that you probably shouldn&#039;t integrate across the non-smooth point at   x  =  0.For instance, for the natural logarithm, logarithms of negative numbers get   i    π added to their positive equivalent. Therefore, if, between 0 (the non-smooth point) and   −  ∞, the constant of integration included    −  i    π, then the two definitions would give you the same result.

I was given a problem to evaluate the following integral, <munderover> ∫

Answered question

Answer & Explanation

New Questions in Pre-Algebra