# Using a function without specifying it

I want to ask sagemath if

$\int_{-\infty}^\infty |x - a| f(x) dx = \int_{-\infty}^a (a -x) f(x) dx + \int_{a}^\infty (x -a) f(x) dx$

which is obviously true without specifying $f(x)$ (a probability density). Is this possible in SageMath ?

I would assume the left-hand side is computed via

but it somehow fails.

Sorry there was a little mistake. I correct it in the text

According to your remark the question is why SageMath is not able return

`true`

to`bool(definite_integral(abs_symbolic(x-a)*f(x), x, -oo, +oo)==definite_integral(abs_symbolic(a-x)*f(x), x, -oo, a)+definite_integral(abs_symbolic(x-a)*f(x), x, a, +oo))`

.One could imagine rewriting the left-hand side by substituting $\lvert x-a \rvert = (x-a)\cdot \chi_{\geqslant a}(x) + (a-x)\cdot \chi_{\lt a}(x)$ where the $\chi$'s are indicator functions, expanding the integrand, using linearity of the integral, and replacing the indicator function factors by narrowed integration bounds. That's easier said than done-with-Sage, though.

This is true for a real function of a real variable and for a real ; in this case,$|x-a|$ is of course either $x-a$ or $a-x$ depending on the

signof $x-a$. If either $x$ or $a$ is complex, this is no longer true.Furthermore, in the complex case, you may have discrepancies if the integration paths of the left and right hands of your equation loop around different poles.