Thomas' Calculus 13th Edition

$\frac{1}{2}-\frac{\pi}{8}$
$\int^{\pi/4}_0 \int^{\ln \sec v}_0 \int^{2t}_{-\infty} e^x$ dx dt dv =$\int^{\pi/4}_0 \int^{\ln \sec v}_0 \lim\limits_{b \to -\infty}(e^{2t}-e^b)$ dt dv =$\int^{\pi/4}_0 \int^{\ln \sec v}_0 e^{2t} dt$ dv =$\int^{\pi/4}_0 (\frac{1}{2}e^{2 \ln \sec v}-\frac{1}{2})dv$ =$\int^{\pi/4}_0 (\frac{sec^2}{v}-\frac{1}{2})dv$ =$[\frac{tan v}{2}-\frac{v}{2}]^{\pi/4}_0$ =$\frac{1}{2}-\frac{\pi}{8}$