Let $f(z)$ be a continuous function on Re$(z)\geq 0$ and suppose that $\lim_{z \rightarrow \infty}f(z) =0$. Then, show that for every negative number $t$
$$ \lim_{R \rightarrow \infty} \int_{\Gamma_{R}} e^{zt}f(z)dz=0, $$
where $\Gamma_{R} = \{|z|=R\} \cap \{Re(z)\geq 0\}.$
My attempt: We know that $|f(z)|$ is a continuous function on $\Gamma_{R}$ so
$$ \| \int_{\Gamma_{R}} e^{zt}f(z)dz \|\leq \int_{\Gamma_{R}} |e^{zt}||f(z)||dz| \leq \sup_{\Gamma_{R}}|f(z)|\int_{\Gamma_{R}} |e^{zt}||dz| $$
Now, if $z=Re^{it}$ we can try to solve the integral, but i don't think this is going to be helpful. I was trying to bound the integral, but i can't figure out how
Partial progress: So all we need to show is that $\int_{\Gamma_{R}} |e^{zt}||dz|$ stays bounded. That integral equals
$$R\int_{-\pi/2}^{\pi/2} e^{tR\cos u}\,du = 2R \int_{0}^{\pi/2} e^{tR\cos u}\,du.$$
Replacing $u$ by $\pi/2 -u$ shows that last integral equals $\int_{0}^{\pi/2} e^{tR\sin u}\,du.$ Now $\sin u \ge (2/\pi)u$ on this interval. It follows that
$$ \int_{0}^{\pi/2} e^{tR\sin u}\,du \le \int_{0}^{\pi/2} e^{tR(2/\pi)u}\,du.$$
I'll stop here. Hopefully this helps.