Krdytkyu

Can someone help with a short algebraic proof that $int_{-a}^ag(x)=int_{-a}^ag(-x)$
From making a sketch this seems to be correct and you could argue from the graph that it would be correct as the function is mirrored in the y-axis. However, I would like to have a short argument that goes better on paper and is more precise. If someone knows where to find an example I would appreciate it a lot.

If someone has another idea the problem arises from proving $frac{1}{2pi}int_{-pi}^{pi}f(x)cdot exp({inx})=frac{1}{2pi}int_{-pi}^{pi}f(-x)cdot exp{(in(-x))}$

Where $i$ is the imaginary number, $n$ is an integer and $f(x)$ is a piecewise differentiable $2pi$ periodic function. This is part of a proof in Fourier analysis.

Let $u=-x$, then:
$$int_{-a}^{a} f(x) mathrm{d}x=int_{a}^{-a} f(-u)mathrm{d}(-u)=-int_a^{-a}f(-u)mathrm{d}u$$
Can you continue?

Every function can be decomposed into the sum of an odd and an even function:
$$f(x)=f_o(x)+f_e(x)$$
Where
$$f_e(x)=frac{f(x)+f(-x)}{2}$$
and
$$f_o(x)=frac{f(x)-f(-x)}{2}$$
So
$$int_{-a}^a f(x)mathrm{d}x = int_{-a}^af_o(x)mathrm{d}x+int_{-a}^af_e(x)mathrm{d}x$$
$$int_{-a}^a f(-x)mathrm{d}x = int_{-a}^af_o(-x)mathrm{d}x+int_{-a}^af_e(-x)mathrm{d}x$$
The integral of an odd function on a symmetric interval is $0$, so we are left with
$$int_{-a}^a f(x)mathrm{d}x = int_{-a}^af_e(x)mathrm{d}x$$
$$int_{-a}^a f(-x)mathrm{d}x = int_{-a}^af_e(-x)mathrm{d}x$$
But $f_e$ is even, so $f_e(x)=f_e(-x)$:
$$int_{-a}^a f(-x)mathrm{d}x = int_{-a}^af_e(-x)mathrm{d}x=int_{-a}^af_e(x)mathrm{d}x=int_{-a}^a f(x)mathrm{d}x$$