Answer
$$\dfrac{1}{2}-\dfrac{\pi}{8}$$
Work Step by Step
Our aim is to integrate the integral as follows:
$$\int^{\pi/4}_0 \int^{\ln \sec v}_0 \int^{2t}_{-\infty} e^x \space dx \space dt \space dv=\int^{\pi/4}_0 \int^{\ln (\sec v)}_0 \lim\limits_{b \to -\infty}(e^{2t}-e^b) \space dt dv \\=\int^{\pi/4}_0 \int^{\ln \sec v}_0 e^{2t} dt \space dv\\=\int^{\pi/4}_0 (\dfrac{1}{2}e^{2 \ln \sec v}-\dfrac{1}{2})dv \\=\int^{\pi/4}_0 (\dfrac{sec^2}{v}-\dfrac{1}{2})dv \\ [\dfrac{\tan (v) }{2}-\dfrac{v}{2}]^{\pi/4}_0 \\=\dfrac{1}{2}-\dfrac{\pi}{8}$$
