Tuesday, June 27, 2017

integration - Why does int+inftyinftyarctanleft(frac11+x2right)dx have a real value when the indefinite integral uses i?




WolframAlpha gives a real closed form for this definite integral:
+arctan(11+x2)dx=2(21)π



Yet, the formula it gives for the indefinite integral uses i.



tan1(1x2+1)dx=xtan1(1x2+1)+2(tan1(x1i)(1i)3/2+tan1(x1+i)(1+i)3/2)+C




Why isn't the definite integral non-real?



Answer



First of all, it is real because it converges and because arctan(11+x2) is a real number for every xR.



On the other hand, that some number can be represented with an expression involving i doesn't mean it is not a real number. For instance,

1i+i=0R.


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