19. $\int_{-\infty}^{0} z e^{2 z} d z$

Step 1 ：First, we recognize that the given integral is an improper integral, so we need to find the limit of the integral as the lower bound approaches negative infinity.

Step 2 ：Let's rewrite the integral as a limit: \(\int_{-\infty}^0 z e^{2z} dz = \lim_{a \to -\infty} \int_{a}^0 z e^{2z} dz\)

Step 3 ：Now, we can use integration by parts to evaluate the integral. Let \(u = z\) and \(dv = e^{2z} dz\). Then, \(du = dz\) and \(v = \frac{1}{2} e^{2z}\).

Step 4 ：Applying integration by parts, we get: \(\int_{a}^0 z e^{2z} dz = \left[ \frac{1}{2} z e^{2z} \right]_a^0 - \int_{a}^0 \frac{1}{2} e^{2z} dz\)

Step 5 ：Now, we can evaluate the remaining integral: \(\int_{a}^0 \frac{1}{2} e^{2z} dz = \left[ -\frac{1}{4} e^{2z} \right]_a^0\)

Step 6 ：Substituting the bounds, we get: \(\left[ \frac{1}{2} z e^{2z} \right]_a^0 - \left[ -\frac{1}{4} e^{2z} \right]_a^0 = \frac{1}{2}(0) e^{2(0)} - \frac{1}{2}(a) e^{2(a)} - \left(-\frac{1}{4} e^{2(0)} + \frac{1}{4} e^{2(a)}\right)\)

Step 7 ：Simplifying, we get: \(\frac{1}{2} - \frac{1}{2} a e^{2a} + \frac{1}{4} - \frac{1}{4} e^{2a}\)

Step 8 ：Now, we need to find the limit as \(a \to -\infty\): \(\lim_{a \to -\infty} \left(\frac{1}{2} - \frac{1}{2} a e^{2a} + \frac{1}{4} - \frac{1}{4} e^{2a}\right)\)

Step 9 ：As \(a \to -\infty\), \(e^{2a} \to 0\), so the limit becomes: \(\lim_{a \to -\infty} \left(\frac{1}{2} + \frac{1}{4}\right) = \frac{3}{4}\)

Step 10 ：Therefore, the value of the integral is \(\boxed{\frac{3}{4}}\)