Answer
Convergent $\;,\;$ $\Large\frac{-1}{4}$
Work Step by Step
Let
\[I=\int_{-\infty}^{0}ze^{2z}dz\]
\[I=\lim_{t\rightarrow -\infty}\int_{t}^{0}ze^{2z}dz\;\;\;\ldots (1)\]
Let \[I_1=\int ze^{2z}dz\]
Using integration by parts
\[I_1=z\int e^{2z}dz-\int \left((z)'\int e^{2z}dz\right)dz\]
\[I_1=\frac{ze^{2z}}{2}-\frac{1}{2}\int e^{2z}dz\]
\[I_1=\frac{ze^{2z}}{2}-\frac{e^{2z}}{4}\;\;\;\ldots (2)\]
Using (2) in (1)
\[I=\lim_{t\rightarrow -\infty}\left[\frac{ze^{2z}}{2}-\frac{e^{2z}}{4}\right]_{t}^{0}\]
\[I=\lim_{t\rightarrow -\infty}\left[\left(0-\frac{1}{4}\right)-\left(\frac{te^{2t}}{2}-\frac{e^{2t}}{4}\right)\right]\]
\[I=\frac{-1}{4}\]
Since limit on R.H.S. of (1) exists so $I$ is convergent
\[I=\frac{-1}{4}.\]
