Question

Suppose that g is an integrable periodic function with period T. Show that for any values of a and b that the definite integral from a to a+T of g(x) is equal to the definite integral from b to b+T of g(x)

Answer 1

\[\int\limits_{a}^{a+T}g(x)dx = \int\limits_{b}^{b+T}g(x)dx\]

Answer 2

alright, here we go~
WLOG, assume a < b
\[\int\limits_{a}^{a+T}g(x)dx = \int\limits_{a}^{b+T}g(x)dx - \int\limits_{a+T}^{b+T}g(x)dx = \int\limits_{a}^{b}g(x)dx+\int\limits_{b}^{b+T}g(x)dx-\int\limits_{a+T}^{b+T}g(x)dx\]
but, because the function is periodic:
\[\int\limits_{a+T}^{b+T}g(x)dx = \int\limits_{a}^{b}g(x)dx\]
So now we have:
\[\int\limits_{a}^{b}g(x)dx+\int\limits_{b}^{b+T}g(x)dx-\int\limits_{a}^{b}g(x)dx = \int\limits_{b}^{b+T}g(x)dx\]

Answer 3

whoops, that should be:
\[\int\limits_{a+T}^{b+T}g(x)dx\]
in the second equation line.

Answer 4

what!? why wont it put my a in the bottom limit lol <.<

Answer 5

you got this. i knew it!

Answer 6

one more time:
\[\int\limits_{a+T}^{b+T}g(x)dx\]

Answer 7

<.< its not me!

Answer 8

\[\int_{a+T}^{b+T}g(x)dx\]

Answer 9

the hell >.>

Answer 10

is that what you want? i have only done this using
\[g(x+t)\]

Answer 11

lolol, well, thats what should be in my second equation line, what satellite posted.

Answer 12

yes thats what i wanted lol

Answer 13

Thanks, so I am guessing I just need to show this for a = b and a > b using similar kinds of arguments?

Answer 14

Whoops scratch a = b....

Answer 15

really scratch that!

Answer 16

and no you don't have to do it again. that is what wlog means

Answer 17

WLOG?

Answer 18

without loss of generality. one of them has to be smaller, so it might as well be a.

Answer 19

oh gotcha thx