We have #g:RR->RR# a continous function with property that #int_(-x)^xg(t)dt=0# for each #x inRR#.How to demonstrate that g is an odd function?

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Answer 1 · 2017-04-20T14:52:53

Define #f(x) = int_0^x g(t) dt# and use the fact that if #f# is even then #f'# is odd.

Suppose that #g# is continuous on #RR# and for every real #x#, we have #int_-x^x g(t) dt = 0#.

Define #f(x) = int_0^x g(t) dt#.

Then for every #x#, #f'(x) = g(x)#

Then #int_-x^x g(t) dt = [ f(t) ]_-x^x = f(x) - f(-x) = 0#

Therefore, #f(-x) = f(x)# so #f# is even.

This implies that #f'# is odd, and since #f' = g#, we see that #g# is odd.

To see that #f# even implies #f'# odd, apply the chain rule to #f(-x)#

Answer 2 · 2017-04-20T15:05:15

See below.

#int_(-x)^x g(t)dt=int_(-x)^x g(-t)(-dt)=-int_(-x)^xg(-t)dt#

so

#int_(-x)^x (g(t)+g(-t))dt=0 forall x# then #g(t) = -g(-t)# so #g(t)# is odd.