Integral of an odd function over a symmetric interval

[sandra1]

Homework Statement

f: [-a,a] >. R is Riemann integrable, prove that ∫[-a, a] ƒ (x) dx = 0

Homework Equations

The Attempt at a Solution

This only proof below I can think of is rather very calculus-ish.I wonder is there any other proof that is more Real Analysis level for this problem? Thanks alot.

since f is an odd function >> -f (x) = f (-x)

∫[-a, a] ƒ (x) dx = ∫[-a, 0] ƒ (x) dx + ∫[0, a] ƒ (x) dx
substitute t = -x with t is dummy variable.
∫[-a, a] ƒ (x) dx = ∫[a, 0] ƒ (-t) -dt + ∫[0, a] ƒ (x) dx (x = -a >> t = a, x = 0 >> t = 0)
= -∫[a, 0] ƒ (-t) dt + ∫[0,a] ƒ (x) dx = ∫[0, a] ƒ (-t) dt + ∫[0,a] ƒ (x) dx
t is dummy is we can rewrite as : = ∫[0,a] ƒ(-x) dx + ∫[0,a] ƒ(x) dx
= ∫[0, a] - ƒ (x) dt +∫[0, a] ƒ (x) dx
(by definition of odd function)
= -∫[0, a] ƒ (x) dt + ∫[0, a] ƒ (x) dx = 0

 

[tiny-tim]

hi sandra1!

This only proof below I can think of is rather very calculus-ish.

yes, particularly the "dummy variable"!

try defining a function g with g(x) = f(-x) ​

 

[Office_Shredder]

Alternatively, if you take a Riemann sum with evenly sized sub-intervals, the oddness of the function ends up canceling almost every term and it's not that hard to show that the integral must be zero (something to think about real quick: why doesn't this prove every odd function is integrable over symmetric intervals?)