Does #int fg = int f int g# ?

Question

Emily Ammerman · Accepted Answer

Proof by contradiction

This is an interesting question, but can be simply prooved via contradiction: Let #int f(x) g(x) dx = int f(x) dx int g(x) dx # This is the statement we are trying to disprove: Let #f(x) = sinx # and #g(x) = cosx # #L.H.S:# # => int sinx cosx dx # #=> 1/2 int sin2x dx # #=> 1/2 * [ -1/2 * cos 2x + c_0] # #= -1/4 cos 2x + c_1 , c_1 in RR# #R.H.S:# #=> int cosx dx int sinx dx# # => (sinx + c_2) * (-cosx + c_3)# # = -sinxcosx + c_3sinx -c_2 cosx + c_2c_3# # = c_2c_3 + c_3sinx - c_2cosx - 1/2 sin2x # # c_2 ,c_3 in RR# # c_2c_3 + c_3sinx - c_2cosx - 1/2 sin2x != -1/4 cos 2x + c_1# Hence we see that #L.H.S != R.H.S # => # int cosxsinx dx != int cosx dx int sinx dx# Hence we were able to select #f(x)# and #g(x)# that didnt hold the statement, we can try this for many other combinations, ie, #f(x) = g(x) = x, # there are infinitely many functions that dont hold #=># Hence proven by contradiction that the statement doesnt hold for all functions #f(x)# and #g(x)# but thats not saying there arent any functions that do hold!

Isaiah Allen · Answer

See explanation...

There are some functions for which: #int fg = int f int g# The easiest examples to find are functions which are mostly #0#. For example: #f(x) = g(x) = { (1 " if " x " is rational"), (0 " if " x " is irrational") :}# Then: #int fg = 0+C = int f int g# If however #f(x)# and #g(x)# are non-zero functions with power series expansions, then we can show: #int fg != int f int g# For example suppose #a_mx^m# is the lowest degree non-zero term of the power series expansion of #f(x)# and #b_nx^n# is the lowest degree non-zero term of the power series expansion of #g(x)#. Then the lowest degree non-zero term of #int f int g# is: #(a_m b_n)/((m+1)(n+1)) x^(m+n+2)# while that of #int fg# is: #(a_m b_n)/(m+n+1) x^(m+n+1)# So: #int fg != int f int g#

Emilia Aguilar · Answer

If at least one of #f#, #g# is a constant function then equality does hold. (Also for the example cited by George C.)

To show that the equality does not hold universally, it suffices to provide a counterexample. That is: to counter the claim that the integrals are equal, we must show that for some #f# and #g#, we have #intfg != intf intg#. (We do not need to show that, for all #f# and #g#, the inequality holds.) Rhys has provided a counterexample using one of my favorite integrals -- #intsinxcosxdx# My favorite counterexample to this "multiplication rule" has always been #f(x) = g(x) =x#. #int (x*x) dx != intx dx int xdx#