Is the Laplace transform of $f(-t)$ given by $F(-s)$?

Question

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 $F(-s)$ ?

Answer 1

Since$$F(-s)=\int_{-\infty}^{+\infty}f(t)e^{ts}\,\mathrm dt,$$if you do $t=-u$ and $\mathrm dt=-\mathrm du$, you get$$F(-s)=-\int_{+\infty}^{-\infty}f(-u)e^{-us}\,\mathrm du=\int_{-\infty}^{+\infty}f(u)e^{-us}\,\mathrm du=F(s).$$

Answer 2

You don't conclude that- it isn't true. For example, take $f(t)= e^t$. Its Laplace transform is $\frac{1}{s- 1}$ but the Laplace transform of $f(-t)= e^{-t}$ is $\frac{1}{s+ 1}$, NOT $\frac{1}{(-s)- 1}$.