Does #int fg = int f int g# ?
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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!
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#
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#
