Suppose $f(t)$ is an *even* function, then we have $f(t) = f(-t)$, thus

$$F(s) = \int\limits\_{-\infty}^{\infty}f(t)e^{-st}\,dt = \int\limits\_{-\infty}^{\infty}f(-t)e^{-st}\,dt $$

From here I'm not so sure how to proceed. I need to somehow show that $F(s)=F(-s)$ when $f(t)$ is an *even* function. To do this, it looks I need to show that the Laplace transform of $f(-t)$ is given by $F(-s)$. Should I substitute $u=-t$ ? Then I get

$$F(s) = \int\limits\_{-\infty}^{\infty}f(t)e^{-st}\,dt = \int\limits\_{-\infty}^{\infty}f(-t)e^{-st} \,dt= \int\limits\_{-\infty}^{\infty}f(u)e^{su}\,dt$$

But this is not in the standard laplace integral form. How to conclude that this equals $F(-s)$ ?