I'm trying to prove to myself that
a0 = $\frac{1}{L}\int_{-L}^Lf$e$(x)dx$ = $\frac{2}{L}\int_{0}^Lf$e$(x)dx$
but I keep getting a0 = 0.
This is my logic at the moment. If:
$f$e$(x)$ = $f(x)$ for $0<x<L$
And
$f$e$(x)$ = $f$e$(-x)$ for all x then:
$\frac{1}{L}\int_{-L}^Lf$e$(x)dx$ = $\frac{1}{L}\int_{-L}^0f$e$(x)dx+\frac{1}{L}\int_{0}^Lf$e(x)dx
= $\frac{1}{L}\int_{L}^0f$e$(-x)dx+\frac{1}{L}\int_{0}^Lf$e(x)dx
= $\frac{1}{L}\int_{L}^0f$e$(x)dx+\frac{1}{L}\int_{0}^Lf$e(x)dx
= $\frac{2}{L}\int_{L}^Lf$e$(x)dx = 0$
I've tried doing this a few different ways and always get this result and I really can't find where I'm going wrong, any and all help appreciated.
Thanks,
Richard
In your next to your last step, it looks like you try to make the substitution
$$x' = -x$$
But you forgot to include a negative sign because
$$dx' = -dx$$