How do you evaluate the integral #int x dx# from #-oo# to #oo# if it converges?

Question

Isaiah Allen · Accepted Answer

#0#

We have:

#int_(-oo)^ooxdx#

Integrating:

#=[x^2/2]_(-oo)^oo#

We can't technically "plug in" infinity and negative infinity, so take their limits:

#=lim_(xrarroo)(x^2/2)-lim_(xrarr-oo)(x^2/2)#

Both approach positive infinity:

#=oo-oo=0#

This should make sense, if we think about #int_(-oo)^ooxdx# as describing the area between the line #y=x# and the #x#-axis. The negative portion where #x<0# is completely balanced out by the positive area where #x>0#:

graph{x [-304.4, 304.4, -152.2, 152.2]}

Christopher Agresta · Answer

undefined If the integral ∫x dx from -∞ to ∞ converges, it can be evaluated using techniques from calculus. Since the integral represents the area under the curve of the function x from negative infinity to positive infinity, it converges if the function x approaches zero sufficiently quickly as x tends to positive or negative infinity. The integral evaluates to zero if the function x is odd, and it diverges if the function x is even. If the integral converges, it can be evaluated using methods such as integration by parts or substitution.