I apologise upfront for the dumb question, but I just totally got stuck with this.
An apparent contradiction when taking the convolution with an even function.
For two (nice) functions $f$ and $g$ consider $$ (f*g)(x)=\int_{-\infty}^{+\infty}f(x-y)g(y) dy. $$
Assume $f$ to be even (i.e., $f(-x)=f(x)$ for all $x$). Pedantically enough, define $$ f_1(x):=f(x),\qquad f_2(x):=f(-x). $$ Clearly, $f_1\equiv f_2$.
Now, $$ \begin{split} &\int_{-\infty}^{+\infty}f(x-y)g(y) dy \\ &= (f*g)(x) = (f_1*g)(x) = (f_2*g)(x) \\ &= \int_{-\infty}^{+\infty}f(-x-y)g(y) \\ &= \int_{-\infty}^{+\infty}f(x+y)g(y) dy. \end{split} $$
Except that the identity $$ \int_{-\infty}^{+\infty}f(x-y)g(y) dy = \int_{-\infty}^{+\infty}f(x+y)g(y) dy $$ is false in general. For instance, with $f(x)=e^{-x^2}$ and $g(x)= \sqrt{\frac{8}{\pi}}x e^{-x^2}$ one has $$ \begin{split} \int_{-\infty}^{+\infty}f(x-y)g(y) dy = x e^{-\frac{x^2}{2}} \\ \int_{-\infty}^{+\infty}f(x+y)g(y) dy = -x e^{-\frac{x^2}{2}}. \end{split} $$
So what went wrong.....?