Answer
Convergent $\;,\;$ $\Large\frac{-π}{8}$
Work Step by Step
Let \[I=\int_{-\infty}^{0}\frac{z}{z^4+4}dz\]
\[I=\lim_{t\rightarrow -\infty}\int_{t}^{0}\frac{z}{z^4+4}dz\;\;\;\ldots (1)\]
Let \[I_1=\int\frac{z}{z^4+4}dz\]
Substitute $\; z^2=r\;\;\;\ldots (2)$
\[\Rightarrow 2zdz=dr\]
\[I_1=\frac{1}{2}\int\frac{dr}{r^2+4}\]
\[I_1=\frac{1}{2}\times\frac{1}{2}\tan^{-1}\frac{r}{2}\]
From (2)
\[I_1=\frac{1}{4}\tan^{-1}\frac{z^2}{2}\;\;\;\ldots (3)\]
Using (3) in (1)
\[I=\lim_{t\rightarrow -\infty}\left[\frac{1}{4}\tan^{-1}\frac{z^2}{2}\right]_{t}^{0}\]
\[I=\lim_{t\rightarrow -\infty}\left[\frac{1}{4}(0)-\frac{1}{4}\tan^{-1}\frac{t^2}{2}\right]\]
\[I=\frac{-1}{4}\times\frac{π}{2}\]
\[I=\frac{-π}{8}\]
Since limit on R.H.S. of (1) exists so $I$ is convergent
and
\[I=\frac{-π}{8}.\]
