How do you integrate (-infinity to 5) 1/(x^2+1)dx?

CherryB

New member
When I integrate
int (-infinity to 5) 1/(x^2+1)dx by using the limit as t --> -infinity of int (t to 5) 1/(x^2+1)dx, I get negative infinity, which means it is DIVERGENT. (antiderivative of 1/(x^2+1) is arctan x, right? and since arctan t as t--> -infinity is -infinity, I get divergent)

But the answer gives a numerical value for the answer, which means it is convergent... why?
 

JohnF

Member
arctan (- infinity) = - pi/2 = -1.5707

arctan (5) = 1.3734

arctan (5) - arctan (- infinity) = 1.3734 - (- 1.5707) = 2.94
 
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