To evaluate the triple integral of $e^{x + y + z}$ with the given limits:

$\int_0^{\log 2} \int_0^x \int_0^{x-y} e^{x + y + z} \, dz \, dy \, dx$

We can evaluate this step-by-step:

Step 1: Inner Integral (with respect to $z$)

The inner integral is: $\int_0^{x - y} e^{x + y + z} \, dz$ Since $e^{x + y}$ is independent of $z$, it acts as a constant: $e^{x + y} \int_0^{x - y} e^z \, dz$ The integral of $e^z$ is simply $e^z$, so we get: $e^{x + y} \left[ e^z \right]_0^{x - y} = e^{x + y} \left( e^{x - y} - e^0 \right) = e^{x + y} \left( e^{x - y} - 1 \right)$ Thus, the result of the inner integral is: $e^{x + y} (e^{x - y} - 1)$

Step 2: Middle Integral (with respect to $y$)

Now we evaluate the middle integral: $\int_0^x e^{x + y} (e^{x - y} - 1) \, dy$ Distribute $e^{x + y}$ across the terms inside the parentheses: $\int_0^x e^{x + y} e^{x - y} \, dy - \int_0^x e^{x + y} \, dy$ Simplify the exponents: $\int_0^x e^{2x} \, dy - \int_0^x e^{x + y} \, dy$ The first integral is straightforward: $e^{2x} \int_0^x 1 \, dy = e^{2x} \cdot x$ For the second integral: $e^x \int_0^x e^y \, dy = e^x \cdot [e^y]_0^x = e^x (e^x - 1)$ So the middle integral becomes: $e^{2x} x - e^x (e^x - 1) = e^{2x} x - e^{2x} + e^x$

Step 3: Outer Integral (with respect to $x$)

Now, we perform the outer integral: $\int_0^{\log 2} \left( e^{2x} x - e^{2x} + e^x \right) \, dx$

We break this into three separate integrals:

1. $\int_0^{\log 2} e^{2x} x \, dx$
2. $\int_0^{\log 2} e^{2x} \, dx$
3. $\int_0^{\log 2} e^x \, dx$

First Integral:

Use integration by parts for $\int e^{2x} x \, dx$: Let $u = x$ and $dv = e^{2x} dx$, so $du = dx$ and $v = \frac{1}{2} e^{2x}$.

Applying the integration by parts formula $\int u \, dv = uv - \int v \, du$, we get: $\int e^{2x} x \, dx = \frac{x}{2} e^{2x} - \frac{1}{2} \int e^{2x} \, dx = \frac{x}{2} e^{2x} - \frac{1}{4} e^{2x}$ Evaluating from $0$ to $\log 2$: $\left[ \frac{x}{2} e^{2x} - \frac{1}{4} e^{2x} \right]_0^{\log 2}$ At $x = \log 2$: $\frac{\log 2}{2} \cdot 4 - \frac{1}{4} \cdot 4 = 2 \log 2 - 1$ At $x = 0$, both terms vanish, so the result is: $2 \log 2 - 1$

Second Integral:

The second integral is: $\int_0^{\log 2} e^{2x} \, dx$ This is a standard exponential integral: $\frac{1}{2} e^{2x} \Big|_0^{\log 2} = \frac{1}{2} (4 - 1) = \frac{3}{2}$

Third Integral:

The third integral is: $\int_0^{\log 2} e^x \, dx = e^x \Big|_0^{\log 2} = 2 - 1 = 1$

Final Result

Now, summing up all three parts: $(2 \log 2 - 1) - \frac{3}{2} + 1 = 2 \log 2 - \frac{1}{2}$

Thus, the value of the triple integral is: $\boxed{2 \log 2 - \frac{1}{2}}$

Would you like more details or clarification? Here are a few follow-up questions:

1. How does the technique of integration by parts apply in this context?
2. Can this triple integral be evaluated using other methods, such as switching the order of integration?
3. What would the integral look like if the limits were changed, for example, if the outer integral were from 0 to 1?
4. How would the result change if we substituted $x = \log 2$ with another expression?
5. What are the general steps to solving a triple integral with non-constant limits?

Tip: When dealing with integrals with non-constant limits, always carefully examine the limits of integration for each variable before integrating.