PROBLEM 3. Let the function $f:[0,+\infty) \rightarrow \mathbb{R}$, defined by $f(x)=\left\{\begin{array}{cl}x, & \text { if } 0 \leq x \leq 1 \\ x-1, & \text { if } x>1\end{array}\right.$. Calculate $\int_{0}^{1} \frac{f\left(e^{t}\right)}{f\left(e^{-t}\right)} \mathrm{d} t$.
## PROBLEM 3.

(2p) $\quad f\left(e^{t}\right)=\left\{\begin{array}{cc}e^{t}, & \text { if } t \leq 0 \\ e^{t}-1, & \text { if } t>0\end{array} ; f\left(e^{-t}\right)=\left\{\begin{array}{cc}e^{-t}, & \text { if } t \geq 0 \\ e^{-t}-1, & \text { if } t<0\end{array}\right.\right.$

(2p) $\quad$ For $t \in[0,1]$, we have $g(t)=\frac{f\left(e^{t}\right)}{f\left(e^{-t}\right)}=\left\{\begin{array}{cl}1, & t=0 \\ e^{2 t}-e^{t}, & 0<t \leq 1\end{array}\right.$

$(2 p) \quad$ The function $g:[0,1] \rightarrow \mathbb{R}$ is bounded and continuous on $(0,1]$, so it is integrable and $\int_{0}^{1} \frac{f\left(e^{t}\right)}{f\left(e^{-t}\right)} \mathrm{d} t=\int_{0}^{1}\left(e^{2 t}-e^{t}\right) \mathrm{d} t$

$(1 p)$ $\int_{0}^{1}\left(e^{2 t}-e^{t}\right) \mathrm{d} t=\frac{1}{2}(e-1)^{2}$