Given a probability density function $f(x)$ such that $\int_{-\infty}^{\infty}f(x)dx=1$. Do we always have $\int_{-\infty}^{\infty}f(-x)dx=1$?

Question

This question here raises from change of integration in integration.

In integration we learn that we can use Jacobi to deal with integration. For example in one dimension, we have derivative as our Jacobian value. However, in probability class, we also learn that we need some kind of absolute value of the Jacobian. In other words, the absolute value of the derivative in one dimension. May I ask why there needs an absolute here? Is it true that for probability integration in higher dimensions, there would need some 'absolute form' of Jacobian matrix ?

Thanks in advance!

Answer 1

Yes, the integrals are identical $$ \int_{-\infty}^\infty f(-x) \; dx \stackrel{(u \ = \ -x)}{=} \int_{\infty}^{-\infty} -f(u) \; du = \int_{-\infty}^\infty f(u) \; du =1 $$