Let's walk through the evaluation of the integral step by step:

$\int_0^{\pi/6} \int_0^{\ln(\sec v)} \int_{-\infty}^{2t} e^x \, dx \, dt \, dv$

Step 1: Evaluate the innermost integral with respect to $x$

We need to integrate $e^x$ from $-\infty$ to $2t$.

$\int_{-\infty}^{2t} e^x \, dx$

The integral of $e^x$ is $e^x$. Now apply the limits:

$\left[ e^x \right]_{-\infty}^{2t} = e^{2t} - \lim_{x \to -\infty} e^x = e^{2t} - 0 = e^{2t}$

So, the innermost integral becomes:

$\int_{-\infty}^{2t} e^x \, dx = e^{2t}$

Step 2: Evaluate the second integral with respect to $t$

Now we substitute the result from the innermost integral into the next one. We now integrate $e^{2t}$ with respect to $t$ from $0$ to $\ln(\sec v)$.

$\int_0^{\ln(\sec v)} e^{2t} \, dt$

The antiderivative of $e^{2t}$ is:

$\frac{e^{2t}}{2}$

Applying the limits:

$\left[ \frac{e^{2t}}{2} \right]_0^{\ln(\sec v)} = \frac{e^{2 \ln(\sec v)}}{2} - \frac{e^0}{2}$

Using the logarithm property $e^{2 \ln(\sec v)} = (\sec v)^2$:

$= \frac{(\sec v)^2}{2} - \frac{1}{2}$

So, the second integral evaluates to:

$\frac{(\sec v)^2 - 1}{2}$

Step 3: Simplify the expression

Using the trigonometric identity $\sec^2 v - 1 = \tan^2 v$, we simplify the expression:

$\frac{(\sec v)^2 - 1}{2} = \frac{\tan^2 v}{2}$

Step 4: Evaluate the outermost integral with respect to $v$

Now we integrate $\frac{\tan^2 v}{2}$ with respect to $v$ from $0$ to $\pi/6$.

$\int_0^{\pi/6} \frac{\tan^2 v}{2} \, dv = \frac{1}{2} \int_0^{\pi/6} \tan^2 v \, dv$

We know that:

$\int \tan^2 v \, dv = \tan v - v$

Applying the limits:

$\left[ \tan v - v \right]_0^{\pi/6} = \left( \tan\left(\frac{\pi}{6}\right) - \frac{\pi}{6} \right) - (0 - 0)$

Since $\tan(\pi/6) = \frac{1}{\sqrt{3}}$, the result becomes:

$\frac{1}{\sqrt{3}} - \frac{\pi}{6}$

Step 5: Final result

Now multiply by the constant $\frac{1}{2}$:

$\frac{1}{2} \left( \frac{1}{\sqrt{3}} - \frac{\pi}{6} \right) = \frac{1}{2\sqrt{3}} - \frac{\pi}{12}$

Thus, the value of the integral is:

$\boxed{\frac{1}{2\sqrt{3}} - \frac{\pi}{12}}$

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Do you have any questions or want further clarification?
Here are five related questions you could explore next:

1. How do we compute improper integrals involving infinity?
2. What are other identities involving $\sec v$ and $\tan v$?
3. How do logarithmic functions behave inside integrals?
4. What are the properties of definite integrals over trigonometric functions?
5. How does integration by parts work for nested integrals?

Tip: When integrating exponential functions over infinite limits, remember that their convergence depends on the sign of the exponent!