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 $\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)\cdot \exp({inx})=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(-x)\cdot \exp{(in(-x))}$
Where $i$ is the imaginary number, $n$ is an integer and $f(x)$ is a piecewise differentiable $2\pi$ 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 $f_e(x)=f_e(-x)$: $$\int_{-a}^a f(-x)\mathrm{d}x = \int_{-a}^af_e(-x)\mathrm{d}x=\int_{-a}^af_e(x)\mathrm{d}x=\int_{-a}^a f(x)\mathrm{d}x$$